CHEM 132 - ADVANCED GENERAL
CHEMISTRY II - Radioactive decay
calculations
Question Bank - Set 5
Liberty University
Question 1
Question
A sample of a radioactive isotope decays according to the formula N(t) =
N0e−kt, where N(t) is the quantity of the isotope remaining after tyears, N0is
the initial quantity of the isotope, and kis a constant. Given that the half-life
of the isotope is 100 years, find the decay constant k.
Solution
Step 1: Recall that the half-life of a substance is the time it takes for half of the
original sample to decay. In this case, the half-life is given as 100 years which
means that after 100 years, N(100) = 1
2N0.
Step 2: Substituting t= 100 and N(t) = 1
2N0into the decay formula, we
get: 1
2N0=N0e−100k
Step 3: Divide both sides by N0to simplify the equation:
1
2=e−100k
Step 4: Take the natural logarithm of both sides to solve for k:
ln 1
2= lne−100k
Step 5: Recall that ln(ex) = x, so we have:
ln 1
2=−100k
Step 6: Solve for kby dividing by −100:
k=ln 1
2
−100
Step 7: Finally, calculate the value of kusing a calculator to get:
k≈ln(0.5)
−100 ≈0.00693 years−1
Therefore, the decay constant kfor the radioactive isotope is approximately
0.00693 years−1.
Question 2
Question
A radioactive substance has a half-life of 10 days. If the initial quantity of the
substance is 500 grams, determine the amount of the substance remaining after
30 days.
Solution
Step 1: Calculate the decay constant (λ) using the formula N(t) = N0·e−λt.
Here, N(t) is the amount of the substance remaining after time t,N0is the
initial quantity, λis the decay constant, and tis the time elapsed.
0.5N0=N0·e−λ(10)
e−10λ= 0.5
−10λ= ln(0.5)
λ=ln(0.5)
−10 =ln2−1
−10 =−ln(2)
10
Step 2: Substitute the decay constant back into the formula to find the
amount of substance remaining after 30 days.
N(30) = 500 ·e−
−ln(2)
10 ·30
N(30) = 500 ·eln(2)·3
N(30) = 500 ·eln(8)
N(30) = 500 ·8
N(30) = 4000 grams
Therefore, the amount of the substance remaining after 30 days is 4000
grams.
2
Question 3
Question
A sample of a radioactive isotope has an initial activity of 5000 decays per
second. After 3 hours, the activity of the sample has decreased to 1250 decays
per second. Calculate the half-life of the isotope.
Solution
Step 1: Find the decay constant, λ, using the decay formula A=A0e−λt, where
Ais the activity at time t,A0is the initial activity, and λis the decay constant.
Given that A0= 5000 decays/sec, A= 1250 decays/sec, and t= 3 hours, we
can rearrange the formula to solve for λ:
1250 = 5000e−λ·3·3600
1250
5000 =e−10800λ
0.25 = e−10800λ
ln(0.25) = −10800λ
λ=ln(0.25)
−10800
Step 2: Calculate the half-life, T1/2, using the relation λ=ln(2)
T1/2. Given that
λ=ln(0.25)
−10800 , we can solve for T1/2:
ln(0.25)
−10800 =ln(2)
T1/2
T1/2=ln(2)
ln(0.25)
−10800
T1/2=ln(2) · −10800
ln(0.25)
T1/2≈2592 hours
Therefore, the half-life of the radioactive isotope is approximately 2592
hours.
Question 4
Question
A certain radioactive substance has a half-life of 500 years. If there are initially
100 grams of the substance, how many grams will remain after 1500 years?
3
Solution
Step 1: Determine the decay constant using the formula: k=ln(2)
T1
2
, where ln
represents the natural logarithm, T1
2is the half-life of the substance, and kis
the decay constant.
Step 1: k=ln(2)
500 ≈0.0013863 years−1
Step 2: Calculate the amount of substance remaining after 1500 years using
the formula: N(t) = N0e−kt, where N(t) is the amount of substance remaining
after time t,N0is the initial amount of substance, kis the decay constant, and
tis the time.
Step 2: N(1500) = 100 ×e−0.0013863×1500 ≈60.653 grams
Answer
After 1500 years, approximately 60.653 grams of the substance will remain.
Question 5
Question
A certain radioactive substance decays according to the equation N(t) = N0e−λt,
where N(t) is the amount of the substance at time t,N0is the initial amount
of the substance, λis the decay constant, and tis the time elapsed.
Given that the initial amount of the substance is 100 grams, the half-life of
the substance is 10 days, and tis 30 days, find the amount of the substance
remaining after 30 days.
Solution
Step 1: Find the decay constant λusing the half-life formula T1/2=ln(2)
λ.
Given T1/2= 10 days
λ=ln(2)
T1/2
=ln(2)
10 ≈0.0693 days−1.
Step 2: Substitute the values into the decay equation N(t) = N0e−λt.
Given N0= 100 grams, t= 30 days
N(30) = 100 ·e−0.0693·30 ≈48.07 grams.
Therefore, the amount of the substance remaining after 30 days is approxi-
mately 48.07 grams.
4
Question 6
Question
A sample of a radioactive material has a half-life of 3 days. If the initial mass
of the sample is 100 grams, determine the mass of the sample after 9 days.
Solution
Step 1: Calculate the decay constant (λ) using the formula T1
2=ln(2)
λ, where
T1
2is the half-life (3 days).
Step 1:
T1
2= 3 days
λ=ln(2)
3= 0.231 days−1
Step 2: Use the exponential decay formula N(t) = N0·e−λt to find the mass
of the sample after 9 days, where N(t) is the final mass, N0is the initial mass,
λis the decay constant, and tis the time passed.
Step 2:
N0= 100 grams
t= 9 days
N(t) = 100 ·e−0.231·9≈30.49 grams
Therefore, the mass of the sample after 9 days is approximately 30.49 grams.
Question 7
Question
A sample of a radioactive material has an initial mass of 10 grams and a half-life
of 3 days. After 9 days, what is the mass of the sample remaining?
Solution
Step 1: Determine the decay constant kusing the half-life formula.
Half-life (T) = ln(2)
k
3 = ln(2)
k
k=ln(2)
3
5
Step 2: Use the decay equation to find the mass of the sample remaining
after 9 days.
m(t) = m0·e−kt
Where: - m(t) is the mass of the sample after time t, - m0is the initial mass of
the sample, - kis the decay constant, - tis the time.
Substitute the given values into the equation:
m(9) = 10 ·e
−
ln(2)
3
(9)
Step 3: Calculate the mass of the sample remaining after 9 days.
m(9) = 10 ·e−3·ln(2)
m(9) = 10 ·eln(2−3)
m(9) = 10 ·eln(1
8)
m(9) = 10 ·1
8
m(9) = 1.25 grams
Therefore, after 9 days, the mass of the sample remaining is 1.25 grams.
Question 8
Question
A sample of a radioactive substance has an initial mass of 500 grams. The
substance decays in such a way that the mass decreases by 201. Determine the
mass of the substance after 15 hours. 2. Calculate the half-life of the radioactive
substance.
Solution
1. To determine the mass of the substance after 15 hours, we first need to find
the decay constant kfrom the given information. Let M(t) be the mass of the
substance after time t. We know that the mass decreases by 20
e−5k= 0.8
Solving for k, we have:
−5k= ln(0.8)
k=−ln(0.8)
5
6
Step 1: Calculate the decay constant k.
k=−ln(0.8)
5≈0.01693
Now, we can find the mass of the substance after 15 hours using the expo-
nential decay formula:
M(15) = M(0) ·ekt
M(15) = 500 ·e0.01693·15
Step 2: Calculate the mass of the substance after 15 hours
M(15) = 500 ·e0.01693·15 ≈500 ·e0.25395
M(15) ≈500 ·1.2899
M(15) ≈644.95 grams
Therefore, the mass of the substance after 15 hours is approximately 644.95
grams.
2. The half-life of a radioactive substance is the time it takes for the sub-
stance to decay to half of its original mass. Let T1
2be the half-life of the
substance. We know that after one half-life, the mass decreases by 50
e−T1
2k= 0.5
Solving for T1
2, we get:
−T1
2k= ln(0.5)
T1
2=−ln(0.5)
k
Step 3: Calculate the half-life of the radioactive substance
T1
2=−ln(0.5)
0.01693 ≈41.05
Therefore, the half-life of the radioactive substance is approximately 41.05
hours.
Question 9
Question
A certain radioactive isotope has a half-life of 5 days. If we start with a sample
containing 1 ×1012 atoms of this isotope, how many atoms will remain after 20
days?
7
Solution
Let’s denote N(t) as the number of remaining atoms of the radioactive isotope
after time t, and N0as the initial number of atoms in the sample.
Step 1: Determine the decay constant λ. Given that the half-life of the
isotope is 5 days, we can use the formula:
λ=ln(2)
T1/2
where ln(2) ≈0.693 is the natural logarithm of 2, and T1/2= 5 days is the
half-life.
λ=0.693
5
λ= 0.1386 days−1
Step 2: Calculate the number of remaining atoms after 20 days using the
radioactive decay formula:
N(t) = N0·e−λt
Given N0= 1 ×1012 atoms and t= 20 days:
N(20) = 1 ×1012 ·e−0.1386·20
N(20) = 1 ×1012 ·e−2.772
N(20) ≈1×1012 ·0.063
N(20) ≈6.3×1010 atoms
Therefore, after 20 days, there will be approximately 6.3×1010 atoms of the
radioactive isotope remaining.
Question 10
Question
A sample of a radioactive substance decays according to the equation N(t) =
N0e−0.05t, where N(t) represents the amount of the substance remaining after
tyears and N0is the initial amount. If the initial amount of the substance is
100 grams, find the amount of substance remaining after 20 years. Round your
answer to the nearest whole number.
Solution
Step 1: Substitute the given values into the equation.
Given: N0= 100 grams
t= 20 years
8
The equation for radioactive decay is N(t) = N0e−0.05t.
Substitute N0= 100 and t= 20 into the equation:
N(20) = 100e−0.05×20
Step 2: Calculate the amount of substance remaining after 20 years.
N(20) = 100e−1≈100 ×0.3679 ≈36.79
Step 3: Round the answer to the nearest whole number.
Rounding 36.79 to the nearest whole number, we get ≈37 grams.
Therefore, the amount of substance remaining after 20 years is approximately
37 grams.
Question 11
Question
A certain radioactive substance has a half-life of 5 hours. If initially there are
100 grams of the substance, how much will be left after 15 hours?
Solution
Step 1: The general formula for radioactive decay is given by N(t) = N0·
(1/2)t/h, where: - N(t) is the quantity of the substance remaining after time t,
-N0is the initial quantity of the substance, - his the half-life of the substance.
Step 2: We are given that the half-life of the substance is 5 hours, N0= 100,
and t= 15 hours. Therefore, we can substitute these values into the formula:
N(15) = 100 ·(1/2)15/5
Step 3: Simplifying the formula, we get:
N(15) = 100 ·(1/2)3= 100 ·1/8 = 12.5
Step 4: Therefore, after 15 hours, there will be 12.5 grams of the substance
remaining.
Question 12
Question
A certain radioactive substance has a half-life of 24 hours. If you start with a
sample of 100 grams, how many grams will remain after 3 days?
9
Solution
Step 1: Calculate the number of half-lives in 3 days. To find how many half-lives
have occurred in 3 days, we divide the total time (3 days) by the half-life of the
substance (24 hours):
3 days ×24 hours
1 day = 72 hours
Number of half-lives = 72 hours
24 hours/half-life = 3 half-lives
Step 2: Calculate the amount of substance remaining after 3 days. The
amount of substance remaining after nhalf-lives can be calculated using the
formula:
Amount remaining = 1
2n
×Initial amount
Substitute n= 3 and the initial amount = 100 grams into the formula:
Amount remaining = 1
23
×100 grams = 1
8×100 = 12.5 grams
Therefore, after 3 days, there will be 12.5 grams of the radioactive substance
remaining.
Question 13
Question
A sample of a radioactive isotope has an initial mass of 200 grams. After 8 hours,
only 25 grams of the original isotope remains. If the half-life of the isotope is
10 hours, determine the decay constant and the age of the sample.
Solution
Step 1: Calculate the decay constant. Step 2: Determine the age of the sample.
0.0.1 Step 1:
The decay constant, λ, can be determined using the formula:
N(t) = N0e−λt
where: N(t) = Final mass, N0= Initial mass, λ= Decay constant, t= Time.
Given that N(t) = 25 grams, N0= 200 grams, and t= 8 hours, we can
rearrange the formula to solve for λ:
25 = 200e−λ·8
25
200 =e−8λ
10
0.125 = e−8λ
Taking the natural logarithm of both sides:
ln(0.125) = lne−8λ
ln(0.125) = −8λ
λ=−ln(0.125)
8
λ≈0.0866 hour−1
Therefore, the decay constant is approximately 0.0866 hour−1.
0.0.2 Step 2:
To find the age of the sample, we use the formula:
λ=ln(2)
T1
2
where: ln(2) is the natural logarithm of 2, T1
2is the half-life.
Given that the half-life is 10 hours, we can calculate λ:
0.0866 = ln(2)
10
ln(2) = 0.0866 ×10
ln(2) = 0.866
The age of the sample, denoted by t, can be calculated using the formula:
t=
ln N0
N(t)
λ
Substitute the known values into the formula:
t=ln 200
25
0.0866
Calculating the age of the sample gives:
t=ln(8)
0.0866 ≈22.6 hours
Therefore, the age of the sample is approximately 22.6 hours.
11
Question 14
Question
A certain radioactive substance decays according to the function N(t) = N0·
e−kt, where N(t) is the amount of substance after time t,N0is the initial amount
of substance, and kis a positive constant. Suppose that 80% of the substance
decays in 10 days. Find the value of kfor this substance.
Solution
Step 1: We are given that 80% of the substance decays in 10 days, meaning that
N(10) = 0.2N0. Since N(t) = N0·e−kt, we have:
0.2N0=N0·e−10k
Step 2: Simplifying the equation, we can cancel out N0from both sides:
0.2 = e−10k
Step 3: Taking the natural logarithm of both sides, we get:
ln(0.2) = −10k
Step 4: Solving for k, we have:
k=−ln(0.2)
10 ≈0.0798
Therefore, the value of kfor this substance is approximately 0.0798.
Question 15
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 3
hours, only 1 gram of the substance remains. If the half-life of the substance is
2 hours, what is the decay constant for this radioactive substance?
Solution
Step 1: Determine the fraction of the substance remaining after 3 hours. Let
Nbe the amount remaining after time t,N0be the initial amount, and t1/2be
the half-life. The decay constant λis related to the half-life by the equation:
λ=ln(2)
t1/2
.
12
Given that N0= 10 grams, N= 1 gram, and t= 3 hours, we can use the
formula:
N=N0·e−λt.
Substitute the given values:
1 = 10 ·e−3λ.
Step 2: Solve for the decay constant λ. Divide both sides by 10:
1
10 =e−3λ.
Take the natural logarithm of both sides to solve for λ:
ln 1
10=−3λ.
ln 1
10= lne−3λ.
Apply the properties of logarithms:
ln 1
10=−3λln(e).
ln 1
10=−3λ.
Therefore, the decay constant λis:
λ=−1
3ln(10).
Question 16
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 30
days, the mass of the sample is reduced to 6 grams due to radioactive decay. If
the half-life of the substance is 5 days, determine the decay constant and the
age of the sample.
Solution
Step 1: Determine the decay constant Given that the half-life of the substance
is 5 days, we can use the formula for radioactive decay:
N(t) = N01
2t
T1
2
13
where: - N(t) is the amount of substance remaining after time t, - N0is the
initial amount of substance, - T1
2is the half-life of the substance.
Substitute the known values:
6 = 10 1
230
5
Step 2: Solve for the decay constant Solving the equation gives:
1
26
=1
2t
5
2−6= 2 t
5
−6 = t
5
t=−30
Thus, the decay constant is −30.
Step 3: Calculate the age of the sample To find the age of the sample, we
use the decay constant in the formula:
t=−
ln N(t)
N0
λ
Substitute the known values:
t=−ln 6
10
−30
t=−ln(0.6)
−30
t≈12.7 days
Therefore, the decay constant is -30 and the age of the sample is approxi-
mately 12.7 days.
Question 17
Question
A radioactive substance decays so that its mass is reduced by 30
14
Solution
Step 1: Let’s denote the mass of the substance after thours by m(t).
Step 2: Since the substance reduces by 30
m(t) = 100 ×1−30
100t
15
.
Step 3: Substitute t= 45 into the equation to find the mass of the substance
after 45 hours:
m(45) = 100 ×1−30
10045
15
.
Step 4: Simplify the expression and calculate the mass:
m(45) = 100 ×1−3
103
= 100 ×7
103
= 100 ×343
1000 = 34.3 grams.
Step 5: Therefore, the mass of the substance after 45 hours is 34.3 grams.
Question 18
Question
A sample of radioactive material initially contains 2.5 grams of a certain isotope.
After 6 hours, only 0.3125 grams of the isotope remain. If the half-life of the
isotope is 3 hours, what is the decay constant λof this isotope? (Hint: Use the
formula N(t) = N0e−λt, where N(t) is the amount of the isotope remaining at
time t,N0is the initial amount, and λis the decay constant.)
Solution
Step 1: We can write the equation for the decay of the isotope as:
N(t) = N0e−λt
where N(t) is the amount of the isotope remaining at time t,N0is the initial
amount, and λis the decay constant.
Step 2: We are given that N0= 2.5 grams, N(6) = 0.3125 grams, and the
half-life T1/2= 3 hours.
Step 3: We can determine the initial decay constant using the half-life for-
mula:
λ=ln(2)
T1/2
Step 4: Substitute the known values into the formula to find λ:
λ=ln(2)
3≈0.2310 hours−1
Step 5: Therefore, the decay constant λof this isotope is approximately
0.2310 hours−1.
15
Question 19
Question
A sample of a radioactive isotope has an initial activity of 500 Bq. After 10
hours, the activity of the sample has decreased to 250 Bq. Calculate the half-life
of the isotope.
Solution
Step 1: Let N0be the initial number of radioactive nuclei, Nbe the number of
radioactive nuclei at time t, and T1/2be the half-life of the isotope. We have
the radioactive decay equation:
N=N0·1
2t
T1/2
Step 2: We are given that the initial activity N0= 500 Bq decreases to
N= 250 Bq after 10 hours. Substituting these values into the equation gives:
250 = 500 ·1
210
T1/2
Step 3: Divide both sides by 500 and take the natural logarithm of both
sides to solve for T1/2:
ln 250
500=10
T1/2
·ln(0.5)
Step 4: Simplify the equation:
−ln(2) = 10
T1/2
·ln(0.5)
Step 5: Divide by ln(0.5) to solve for T1/2:
T1/2=10 ·ln(2)
ln(0.5) ≈10 ·0.693
−0.693 ≈ −10 hours
Step 6: Since the half-life cannot be negative, consider the absolute value:
T1/2≈10 hours
Therefore, the half-life of the radioactive isotope is 10 hours.
Question 20
Question
A certain radioactive substance has a half-life of 30 years. If there are initially
100 grams of the substance, how much will remain after 90 years?
16
Solution
Step 1: The formula to calculate the amount of a radioactive substance remain-
ing after a certain time is given by:
N(t) = N01
2t
T1/2
where: - N(t) is the amount of the substance remaining after time t, - N0is the
initial amount of the substance, - T1/2is the half-life of the substance.
Step 2: Substituting the given values into the formula, we get:
N(90) = 100 1
290
30
Step 3: Simplifying the expression, we have:
N(90) = 100 1
23
= 100 1
8=100
8= 12.5 grams
Therefore, after 90 years, there will be 12.5 grams of the radioactive sub-
stance remaining.
Question 21
Question
A certain radioactive substance has a half-life of 10 days. If you start with 100
grams of the substance, how much will remain after 30 days?
Solution
Step 1: Determine the fraction of the substance remaining after 30 days. The
fraction of the substance remaining after a certain amount of time can be cal-
culated using the formula:
Amount remaining = 1
2time
half-life
In this case:
Amount remaining = 1
230
10
=1
23
=1
8
Step 2: Calculate the amount of the substance remaining after 30 days. If
you start with 100 grams of the substance, the amount remaining after 30 days
can be found by multiplying the initial amount by the fraction remaining:
Amount remaining = 100 ×1
8= 12.5 grams
Therefore, after 30 days, there will be 12.5 grams of the radioactive substance
remaining.
17
Question 22
Question
A certain radioactive material decays according to the function A(t) = A0e−kt,
where A(t) is the amount of material at time t,A0is the initial amount of
material, and kis a positive constant. If 80% of the material remains after 10
years, find the value of k.
Solution
Step 1: Given the information, we can set up the equation A(10) = 0.80A0.
A(10) = A0e−10k= 0.80A0
Step 2: Divide by A0to solve for e−10k.
e−10k= 0.80
Step 3: Take the natural logarithm of both sides.
lne−10k= ln(0.80)
Step 4: Use the properties of logarithms to simplify the left side.
−10k= ln(0.80)
Step 5: Divide by −10 to solve for k.
k=ln(0.80)
−10
Step 6: Calculate the value of kusing a calculator.
k≈ln(0.80)
−10 ≈ −0.0721
Therefore, the value of kis approximately −0.0721.
Question 23
Question
A radioactive substance decays according to the equation N(t) = N0e−0.03t,
where N(t) represents the quantity of the substance at time t(in years), N0is
the initial quantity of the substance, and tis the time elapsed since the initial
quantity was measured. If the initial quantity of the substance is 100 grams,
determine how long it will take for the quantity of the substance to reduce to
50 grams.
18
Solution
Step 1: Substitute the given values into the decay equation to find the time it
takes for the substance to reduce to 50 grams.
50 = 100 ·e−0.03t
50
100 =e−0.03t
0.5 = e−0.03t
Step 2: Take the natural logarithm of both sides to solve for t.
ln(0.5) = lne−0.03t
ln(0.5) = −0.03t
t=ln(0.5)
−0.03
t≈23.1 years
Thus, it will take approximately 23.1 years for the quantity of the substance
to reduce to 50 grams.
Question 24
Question
A sample of a radioactive substance decays over time according to the model
Q(t) = Q0·e−kt, where Q(t) represents the amount of the substance remaining
at time t,Q0is the initial amount of the substance, and kis the decay constant.
Given that the initial amount of the substance is 100 grams and after 10 hours
only 30 grams remain, determine the decay constant k.
Solution
Step 1: Given that Q(t) = Q0·e−kt, we can use the information provided at
t= 0 and t= 10 to set up two equations. At t= 0: Q(0) = 100 = Q0·e0=Q0
At t= 10: Q(10) = 30 = 100 ·e−10k
Step 2: Divide the equation at t= 10 by the equation at t= 0 to eliminate
Q0.30
100 =e−10k
Step 3: Simplify the equation.
0.3 = e−10k
Step 4: Take the natural logarithm of both sides to solve for k.
ln(0.3) = lne−10k=−10k
19
Step 5: Solve for k.
k=ln(0.3)
−10 ≈0.1054
Therefore, the decay constant k≈0.1054.
Question 25
Question
A sample of radioactive material decays over time according to the function
N(t) = N0e−kt, where N(t) represents the amount of material left after tyears,
N0is the initial amount of material, and kis a decay constant.
Given that an initial sample of a radioactive material has a half-life of 10
years, find the amount of material left after 30 years.
Solution
Step 1: Find the decay constant kusing the half-life given.
The half-life of a radioactive material is the time it takes for half of the
material to decay. In this case, we have N(t) = N0
2when t= 10 years.
Substitute N(t) = N0
2and t= 10 years into the decay function:
N0
2=N0e−10k
Solving for k:
1
2=e−10k
ln 1
2= lne−10k
ln 1
2=−10k
k=−1
10 ln 1
2
k≈0.0693 (rounded to 4 decimal places)
Step 2: Find the amount of material left after 30 years using the decay
function.
Substitute N0= 1 (let’s assume 1 unit for easier calculation) and t= 30
years into the decay function:
N(30) = 1 ·e−0.0693·30
Calculating N(30):
N(30) = e−2.079
N(30) ≈0.125 (rounded to 3 decimal places)
20
Therefore, after 30 years, approximately 0.125 units of the radioactive ma-
terial remain.
Question 26
Question
The radioactive isotope iodine-131 decays into xenon-131 with a half-life of 8
days. If a sample initially contains 250 grams of iodine-131, how much will
remain after 32 days?
Solution
Step 1: Determine the decay constant. Step 2: Use the decay constant to find
the amount remaining after 32 days.
Step 1: The decay constant λcan be calculated using the formula:
λ=ln(2)
T1/2
where T1/2is the half-life of the isotope. Substitute T1/2= 8 days into the
formula:
λ=ln(2)
8≈0.0866 days−1
Step 2: The amount remaining after time tcan be calculated using the
formula:
N(t) = N0·e−λt
where: - N(t) is the amount remaining after time t, - N0is the initial amount,
-λis the decay constant, and - tis the time in days. We are given N0= 250
grams, t= 32 days, and λ= 0.0866 days−1. Substitute these values into the
formula:
N(32) = 250 ·e−0.0866·32
N(32) = 250 ·e−2.7712
N(32) ≈250 ·0.0625
N(32) ≈15.625 grams
Therefore, approximately 15.625 grams of iodine-131 will remain after 32
days.
21
Question 27
Question
An unknown radioactive substance decays according to the equation A(t) =
A0e−kt where A(t) is the amount of substance at time t(in years), A0is the
initial amount of substance, and kis the decay constant. Given that the half-life
of the substance is 10 years, find the decay constant k.
Solution
Step 1: Recall that the half-life of a radioactive substance is the time it takes
for half of the initial amount to decay. Using the half-life information given, we
can write the following equation:
A0
2=A0e−k·10
Step 2: Divide both sides by A0to simplify the equation:
1
2=e−10k
Step 3: Take the natural logarithm of both sides of the equation to eliminate
the exponential:
ln 1
2= lne−10k
Step 4: Use the property of logarithms that lnab=bln(a) to simplify the
right side:
ln 1
2=−10kln(e)
Step 5: Recall that ln(e) = 1, so the equation simplifies to:
ln 1
2=−10k
Step 6: Solve for the decay constant kby dividing both sides by −10:
k=−ln 1
2
10 =ln(2)
10
Hence, the decay constant kis k=ln(2)
10 .
Question 28
Question
A sample of a radioactive isotope has an initial activity of 2000 decays per
second. After 3 hours, the activity of the sample has decreased to 500 decays
per second. Calculate the half-life of the isotope.
22
Solution
Step 1: Find the decay constant λusing the formula:
N(t) = N0·e−λt
where N(t) is the activity at time t,N0is the initial activity, and tis the time
elapsed.
Given N0= 2000 decays per second, N(t) = 500 decays per second, and
t= 3 hours, we can rearrange the formula to solve for λ:
500 = 2000 ·e−λ·3·3600
500
2000 =e−10800λ
1
4=e−10800λ
Step 2: Solve for the decay constant λ:
ln 1
4=−10800λ
λ=ln(4)
10800 ≈ −2.53 ×10−4s−1
Step 3: Calculate the half-life T1/2using the formula:
T1/2=ln(2)
λ
Substitute the value of λinto the formula:
T1/2=ln(2)
−2.53 ×10−4
T1/2≈2738.7 seconds ≈45.6 minutes
Therefore, the half-life of the isotope is approximately 45.6 minutes.
Question 29
Question
A sample of a radioactive substance decays according to the equation N(t) =
N0·e−0.02t, where N(t) is the number of atoms remaining after tdays, N0is
the initial number of atoms, and tis the time in days. If the initial number of
atoms is 1012, find the rate of decay after 5 days.
23
Solution
Step 1: Calculate the number of atoms remaining after 5 days.
N(5) = N0·e−0.02·5
= 1012 ·e−0.1
≈1012 ·0.9048
≈9.048 ×1011
Step 2: Find the rate of decay after 5 days. The rate of decay is given by
the derivative of N(t) with respect to t, which is dN
dt =−0.02N0·e−0.02t. Now,
substitute t= 5 and N0= 1012 into the equation:
dN
dt t=5
=−0.02 ·1012 ·e−0.02·5
=−0.02 ·1012 ·e−0.1
≈ −0.02 ·1012 ·0.9048
≈ −1.810 ×1011
Therefore, the rate of decay after 5 days is approximately 1.810×1011 atoms
per day.
Question 30
Question
A sample of radioactive element decays according to the equation N(t) =
N0e−0.01t, where N(t) is the amount of the element remaining after tdays,
N0is the initial amount of the element, and tis the time in days. If the ini-
tial amount of the element is 100 grams, what is the amount of the element
remaining after 20 days?
Solution
Step 1: Substitute the given values into the equation to find the amount of the
element remaining.
N(t) = N0e−0.01t
N(20) = 100e−0.01×20
N(20) = 100e−0.2
Step 2: Calculate the amount of the element remaining after 20 days.
N(20) ≈100 ×0.8187
N(20) ≈81.87 grams
Therefore, the amount of the element remaining after 20 days is approxi-
mately 81.87 grams.
24
Question 31
Question
A sample of a radioactive substance has an initial mass of 500 grams. After 30
days, the mass has decreased to 375 grams. If the half-life of the substance is
10 days, what is the decay constant λof the substance?
Solution
Let’s denote the initial mass of the substance as m0= 500 grams, the final
mass as m30 = 375 grams, and the half-life as T1/2= 10 days. We can use the
formula for radioactive decay to determine the decay constant λ.
Step 1: Calculate the fraction of the substance remaining after 30 days.
The fraction of the substance remaining after time tcan be given by mt
m0=e−λt,
where mtis the mass of the substance after time t. Given that m30 = 375 grams
and m0= 500 grams, we have:
375
500 =e−λ×30
Step 2: Calculate the decay constant λ. From Step 1, we have:
3
4=e−30λ
Taking the natural logarithm of both sides gives:
ln 3
4=−30λ
λ=−ln 3
4
30
Step 3: Calculate the decay constant λ.
λ=−ln 3
4
30 ≈0.0513 days−1
Therefore, the decay constant of the substance is approximately 0.0513
days−1.
Question 32
Question
A certain radioactive substance decays at a rate proportional to the amount
of the substance present. Initially, there are 100 grams of the substance, and
10 grams of the substance remain after 20 days. Determine the half-life of the
substance.
25
Solution
Let A(t) denote the amount of the substance present at time t. Since the
substance decays at a rate proportional to its amount, we have the differential
equation dA
dt =−kA,
where kis the decay constant.
Step 1: Solve the differential equation using separation of variables.
ZdA
A=−kZdt =⇒ln |A|=−kt +C,
where Cis the constant of integration.
Step 2: Apply the initial condition, A(0) = 100, to find the value of C.
ln |100|=−k(0) + C=⇒C= ln(100)
Step 3: Use the value of Cto rewrite the equation as A(t) = e−kt+ln(100) =
100e−kt.
Step 4: Apply the second condition that there are 10 grams of the substance
remaining after 20 days to find the value of k.
A(20) = 100e−20k= 10 =⇒e−20k=1
10 =⇒ −20k= ln 1
10
Step 5: Solve for kand then determine the half-life of the substance.
k=−1
20 ln 1
10=ln(10)
20 ≈0.1155
The half-life of the substance is given by T1/2=ln(2)
k.
T1/2=ln(2)
ln(10)
20
=20 ln(2)
ln(10) ≈6.0151 days
Therefore, the half-life of the substance is approximately 6.0151 days.
Question 33
Question
A certain radioactive substance decays according to the formula N(t) = N0e−0.02t,
where N(t) represents the quantity of the substance after tyears and N0is the
initial quantity. If the initial quantity of the substance is 500 grams, determine
the amount of substance remaining after 20 years.
26
Solution
Step 1: Substitute N0= 500 grams and t= 20 years into the given formula.
N(20) = 500e−0.02×20
Step 2: Simplify the expression inside the exponential function.
N(20) = 500e−0.4
Step 3: Using the property of exponents (e−x=1
ex), rewrite the expression.
N(20) = 500
e0.4
Step 4: Calculate the value of e0.4.
e0.4≈1.4918
Step 5: Substitute the value back and calculate N(20).
N(20) = 500
1.4918 ≈335.16 grams
Therefore, the amount of the substance remaining after 20 years is approxi-
mately 335.16 grams.
Question 34
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 500
years, only 2 grams of the substance remain. If the half-life of the substance is
300 years, determine the decay constant and the equation that models the mass
of the substance as a function of time.
Solution
Step 1: Determine the decay constant. Given that the half-life of the substance
is 300 years, we can use the formula for exponential decay to find the decay
constant, denoted by λ. The formula is given by:
N(t) = N0·e−λt,
where N(t) is the remaining mass of the substance at time t,N0is the initial
mass, and λis the decay constant.
We know that N(300) = N0/2 (half-life) and N(500) = 2 grams. Plugging
these values into the formula, we get two equations:
(N0·e−300λ=N0
2
N0·e−500λ= 2
27
Step 2: Solve for the decay constant. By dividing the second equation by
the first, we can eliminate N0:
e−500λ+300λ=2
N0
2
⇒e−200λ= 4 ⇒ −200λ= ln 4 ⇒λ=−ln 4
200
Step 3: Determine the equation modeling the mass. Using the decay constant
found in Step 2, the equation modeling the remaining mass of the substance as
a function of time tis:
N(t) = 10 ·e−ln 4
200 t
Question 35
Question
A certain radioactive isotope decays at a rate proportional to the amount of the
isotope present. The initial amount of the isotope is 100 grams and after 10
days, only 60 grams are left. Calculate the half-life of this isotope.
Solution
Step 1: Let’s denote the half-life of the isotope by T. We know that the amount
of radioactive material Aat time tis given by the formula:
A(t) = A0·e−kt
where A0is the initial amount of the isotope, kis the decay constant, and t
is time.
Step 2: We are given that the initial amount of the isotope is 100 grams,
A0= 100, and after 10 days, only 60 grams are left, A(10) = 60.
Step 3: Substituting the given values into the formula, we have:
60 = 100 ·e−10k
Step 4: Solve for the decay constant k:
e−10k= 0.6
−10k= ln(0.6)
k=−ln(0.6)
10
Step 5: The half-life Tis related to the decay constant kby the formula:
T=ln(2)
k
28
Step 6: Solve for kby dividing by −100:
k=ln 1
2
−100
Step 7: Finally, calculate the value of kusing a calculator to get:
k≈ln(0.5)
−100 ≈0.00693 years−1
Therefore, the decay constant kfor the radioactive isotope is approximately
0.00693 years−1.
Question 2
Question
A radioactive substance has a half-life of 10 days. If the initial quantity of the
substance is 500 grams, determine the amount of the substance remaining after
30 days.
Solution
Step 1: Calculate the decay constant (λ) using the formula N(t) = N0·e−λt.
Here, N(t) is the amount of the substance remaining after time t,N0is the
initial quantity, λis the decay constant, and tis the time elapsed.
0.5N0=N0·e−λ(10)
e−10λ= 0.5
−10λ= ln(0.5)
λ=ln(0.5)
−10 =ln2−1
−10 =−ln(2)
10
Step 2: Substitute the decay constant back into the formula to find the
amount of substance remaining after 30 days.
N(30) = 500 ·e−
−ln(2)
10 ·30
N(30) = 500 ·eln(2)·3
N(30) = 500 ·eln(8)
N(30) = 500 ·8
N(30) = 4000 grams
Therefore, the amount of the substance remaining after 30 days is 4000
grams.
2
Question 3
Question
A sample of a radioactive isotope has an initial activity of 5000 decays per
second. After 3 hours, the activity of the sample has decreased to 1250 decays
per second. Calculate the half-life of the isotope.
Solution
Step 1: Find the decay constant, λ, using the decay formula A=A0e−λt, where
Ais the activity at time t,A0is the initial activity, and λis the decay constant.
Given that A0= 5000 decays/sec, A= 1250 decays/sec, and t= 3 hours, we
can rearrange the formula to solve for λ:
1250 = 5000e−λ·3·3600
1250
5000 =e−10800λ
0.25 = e−10800λ
ln(0.25) = −10800λ
λ=ln(0.25)
−10800
Step 2: Calculate the half-life, T1/2, using the relation λ=ln(2)
T1/2. Given that
λ=ln(0.25)
−10800 , we can solve for T1/2:
ln(0.25)
−10800 =ln(2)
T1/2
T1/2=ln(2)
ln(0.25)
−10800
T1/2=ln(2) · −10800
ln(0.25)
T1/2≈2592 hours
Therefore, the half-life of the radioactive isotope is approximately 2592
hours.
Question 4
Question
A certain radioactive substance has a half-life of 500 years. If there are initially
100 grams of the substance, how many grams will remain after 1500 years?
3
Solution
Step 1: Determine the decay constant using the formula: k=ln(2)
T1
2
, where ln
represents the natural logarithm, T1
2is the half-life of the substance, and kis
the decay constant.
Step 1: k=ln(2)
500 ≈0.0013863 years−1
Step 2: Calculate the amount of substance remaining after 1500 years using
the formula: N(t) = N0e−kt, where N(t) is the amount of substance remaining
after time t,N0is the initial amount of substance, kis the decay constant, and
tis the time.
Step 2: N(1500) = 100 ×e−0.0013863×1500 ≈60.653 grams
Answer
After 1500 years, approximately 60.653 grams of the substance will remain.
Question 5
Question
A certain radioactive substance decays according to the equation N(t) = N0e−λt,
where N(t) is the amount of the substance at time t,N0is the initial amount
of the substance, λis the decay constant, and tis the time elapsed.
Given that the initial amount of the substance is 100 grams, the half-life of
the substance is 10 days, and tis 30 days, find the amount of the substance
remaining after 30 days.
Solution
Step 1: Find the decay constant λusing the half-life formula T1/2=ln(2)
λ.
Given T1/2= 10 days
λ=ln(2)
T1/2
=ln(2)
10 ≈0.0693 days−1.
Step 2: Substitute the values into the decay equation N(t) = N0e−λt.
Given N0= 100 grams, t= 30 days
N(30) = 100 ·e−0.0693·30 ≈48.07 grams.
Therefore, the amount of the substance remaining after 30 days is approxi-
mately 48.07 grams.
4
Question 6
Question
A sample of a radioactive material has a half-life of 3 days. If the initial mass
of the sample is 100 grams, determine the mass of the sample after 9 days.
Solution
Step 1: Calculate the decay constant (λ) using the formula T1
2=ln(2)
λ, where
T1
2is the half-life (3 days).
Step 1:
T1
2= 3 days
λ=ln(2)
3= 0.231 days−1
Step 2: Use the exponential decay formula N(t) = N0·e−λt to find the mass
of the sample after 9 days, where N(t) is the final mass, N0is the initial mass,
λis the decay constant, and tis the time passed.
Step 2:
N0= 100 grams
t= 9 days
N(t) = 100 ·e−0.231·9≈30.49 grams
Therefore, the mass of the sample after 9 days is approximately 30.49 grams.
Question 7
Question
A sample of a radioactive material has an initial mass of 10 grams and a half-life
of 3 days. After 9 days, what is the mass of the sample remaining?
Solution
Step 1: Determine the decay constant kusing the half-life formula.
Half-life (T) = ln(2)
k
3 = ln(2)
k
k=ln(2)
3
5
Step 2: Use the decay equation to find the mass of the sample remaining
after 9 days.
m(t) = m0·e−kt
Where: - m(t) is the mass of the sample after time t, - m0is the initial mass of
the sample, - kis the decay constant, - tis the time.
Substitute the given values into the equation:
m(9) = 10 ·e
−
ln(2)
3
(9)
Step 3: Calculate the mass of the sample remaining after 9 days.
m(9) = 10 ·e−3·ln(2)
m(9) = 10 ·eln(2−3)
m(9) = 10 ·eln(1
8)
m(9) = 10 ·1
8
m(9) = 1.25 grams
Therefore, after 9 days, the mass of the sample remaining is 1.25 grams.
Question 8
Question
A sample of a radioactive substance has an initial mass of 500 grams. The
substance decays in such a way that the mass decreases by 201. Determine the
mass of the substance after 15 hours. 2. Calculate the half-life of the radioactive
substance.
Solution
1. To determine the mass of the substance after 15 hours, we first need to find
the decay constant kfrom the given information. Let M(t) be the mass of the
substance after time t. We know that the mass decreases by 20
e−5k= 0.8
Solving for k, we have:
−5k= ln(0.8)
k=−ln(0.8)
5
6
Step 1: Calculate the decay constant k.
k=−ln(0.8)
5≈0.01693
Now, we can find the mass of the substance after 15 hours using the expo-
nential decay formula:
M(15) = M(0) ·ekt
M(15) = 500 ·e0.01693·15
Step 2: Calculate the mass of the substance after 15 hours
M(15) = 500 ·e0.01693·15 ≈500 ·e0.25395
M(15) ≈500 ·1.2899
M(15) ≈644.95 grams
Therefore, the mass of the substance after 15 hours is approximately 644.95
grams.
2. The half-life of a radioactive substance is the time it takes for the sub-
stance to decay to half of its original mass. Let T1
2be the half-life of the
substance. We know that after one half-life, the mass decreases by 50
e−T1
2k= 0.5
Solving for T1
2, we get:
−T1
2k= ln(0.5)
T1
2=−ln(0.5)
k
Step 3: Calculate the half-life of the radioactive substance
T1
2=−ln(0.5)
0.01693 ≈41.05
Therefore, the half-life of the radioactive substance is approximately 41.05
hours.
Question 9
Question
A certain radioactive isotope has a half-life of 5 days. If we start with a sample
containing 1 ×1012 atoms of this isotope, how many atoms will remain after 20
days?
7
Solution
Let’s denote N(t) as the number of remaining atoms of the radioactive isotope
after time t, and N0as the initial number of atoms in the sample.
Step 1: Determine the decay constant λ. Given that the half-life of the
isotope is 5 days, we can use the formula:
λ=ln(2)
T1/2
where ln(2) ≈0.693 is the natural logarithm of 2, and T1/2= 5 days is the
half-life.
λ=0.693
5
λ= 0.1386 days−1
Step 2: Calculate the number of remaining atoms after 20 days using the
radioactive decay formula:
N(t) = N0·e−λt
Given N0= 1 ×1012 atoms and t= 20 days:
N(20) = 1 ×1012 ·e−0.1386·20
N(20) = 1 ×1012 ·e−2.772
N(20) ≈1×1012 ·0.063
N(20) ≈6.3×1010 atoms
Therefore, after 20 days, there will be approximately 6.3×1010 atoms of the
radioactive isotope remaining.
Question 10
Question
A sample of a radioactive substance decays according to the equation N(t) =
N0e−0.05t, where N(t) represents the amount of the substance remaining after
tyears and N0is the initial amount. If the initial amount of the substance is
100 grams, find the amount of substance remaining after 20 years. Round your
answer to the nearest whole number.
Solution
Step 1: Substitute the given values into the equation.
Given: N0= 100 grams
t= 20 years
8
The equation for radioactive decay is N(t) = N0e−0.05t.
Substitute N0= 100 and t= 20 into the equation:
N(20) = 100e−0.05×20
Step 2: Calculate the amount of substance remaining after 20 years.
N(20) = 100e−1≈100 ×0.3679 ≈36.79
Step 3: Round the answer to the nearest whole number.
Rounding 36.79 to the nearest whole number, we get ≈37 grams.
Therefore, the amount of substance remaining after 20 years is approximately
37 grams.
Question 11
Question
A certain radioactive substance has a half-life of 5 hours. If initially there are
100 grams of the substance, how much will be left after 15 hours?
Solution
Step 1: The general formula for radioactive decay is given by N(t) = N0·
(1/2)t/h, where: - N(t) is the quantity of the substance remaining after time t,
-N0is the initial quantity of the substance, - his the half-life of the substance.
Step 2: We are given that the half-life of the substance is 5 hours, N0= 100,
and t= 15 hours. Therefore, we can substitute these values into the formula:
N(15) = 100 ·(1/2)15/5
Step 3: Simplifying the formula, we get:
N(15) = 100 ·(1/2)3= 100 ·1/8 = 12.5
Step 4: Therefore, after 15 hours, there will be 12.5 grams of the substance
remaining.
Question 12
Question
A certain radioactive substance has a half-life of 24 hours. If you start with a
sample of 100 grams, how many grams will remain after 3 days?
9
Solution
Step 1: Calculate the number of half-lives in 3 days. To find how many half-lives
have occurred in 3 days, we divide the total time (3 days) by the half-life of the
substance (24 hours):
3 days ×24 hours
1 day = 72 hours
Number of half-lives = 72 hours
24 hours/half-life = 3 half-lives
Step 2: Calculate the amount of substance remaining after 3 days. The
amount of substance remaining after nhalf-lives can be calculated using the
formula:
Amount remaining = 1
2n
×Initial amount
Substitute n= 3 and the initial amount = 100 grams into the formula:
Amount remaining = 1
23
×100 grams = 1
8×100 = 12.5 grams
Therefore, after 3 days, there will be 12.5 grams of the radioactive substance
remaining.
Question 13
Question
A sample of a radioactive isotope has an initial mass of 200 grams. After 8 hours,
only 25 grams of the original isotope remains. If the half-life of the isotope is
10 hours, determine the decay constant and the age of the sample.
Solution
Step 1: Calculate the decay constant. Step 2: Determine the age of the sample.
0.0.1 Step 1:
The decay constant, λ, can be determined using the formula:
N(t) = N0e−λt
where: N(t) = Final mass, N0= Initial mass, λ= Decay constant, t= Time.
Given that N(t) = 25 grams, N0= 200 grams, and t= 8 hours, we can
rearrange the formula to solve for λ:
25 = 200e−λ·8
25
200 =e−8λ
10
0.125 = e−8λ
Taking the natural logarithm of both sides:
ln(0.125) = lne−8λ
ln(0.125) = −8λ
λ=−ln(0.125)
8
λ≈0.0866 hour−1
Therefore, the decay constant is approximately 0.0866 hour−1.
0.0.2 Step 2:
To find the age of the sample, we use the formula:
λ=ln(2)
T1
2
where: ln(2) is the natural logarithm of 2, T1
2is the half-life.
Given that the half-life is 10 hours, we can calculate λ:
0.0866 = ln(2)
10
ln(2) = 0.0866 ×10
ln(2) = 0.866
The age of the sample, denoted by t, can be calculated using the formula:
t=
ln N0
N(t)
λ
Substitute the known values into the formula:
t=ln 200
25
0.0866
Calculating the age of the sample gives:
t=ln(8)
0.0866 ≈22.6 hours
Therefore, the age of the sample is approximately 22.6 hours.
11
Question 14
Question
A certain radioactive substance decays according to the function N(t) = N0·
e−kt, where N(t) is the amount of substance after time t,N0is the initial amount
of substance, and kis a positive constant. Suppose that 80% of the substance
decays in 10 days. Find the value of kfor this substance.
Solution
Step 1: We are given that 80% of the substance decays in 10 days, meaning that
N(10) = 0.2N0. Since N(t) = N0·e−kt, we have:
0.2N0=N0·e−10k
Step 2: Simplifying the equation, we can cancel out N0from both sides:
0.2 = e−10k
Step 3: Taking the natural logarithm of both sides, we get:
ln(0.2) = −10k
Step 4: Solving for k, we have:
k=−ln(0.2)
10 ≈0.0798
Therefore, the value of kfor this substance is approximately 0.0798.
Question 15
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 3
hours, only 1 gram of the substance remains. If the half-life of the substance is
2 hours, what is the decay constant for this radioactive substance?
Solution
Step 1: Determine the fraction of the substance remaining after 3 hours. Let
Nbe the amount remaining after time t,N0be the initial amount, and t1/2be
the half-life. The decay constant λis related to the half-life by the equation:
λ=ln(2)
t1/2
.
12
Given that N0= 10 grams, N= 1 gram, and t= 3 hours, we can use the
formula:
N=N0·e−λt.
Substitute the given values:
1 = 10 ·e−3λ.
Step 2: Solve for the decay constant λ. Divide both sides by 10:
1
10 =e−3λ.
Take the natural logarithm of both sides to solve for λ:
ln 1
10=−3λ.
ln 1
10= lne−3λ.
Apply the properties of logarithms:
ln 1
10=−3λln(e).
ln 1
10=−3λ.
Therefore, the decay constant λis:
λ=−1
3ln(10).
Question 16
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 30
days, the mass of the sample is reduced to 6 grams due to radioactive decay. If
the half-life of the substance is 5 days, determine the decay constant and the
age of the sample.
Solution
Step 1: Determine the decay constant Given that the half-life of the substance
is 5 days, we can use the formula for radioactive decay:
N(t) = N01
2t
T1
2
13
where: - N(t) is the amount of substance remaining after time t, - N0is the
initial amount of substance, - T1
2is the half-life of the substance.
Substitute the known values:
6 = 10 1
230
5
Step 2: Solve for the decay constant Solving the equation gives:
1
26
=1
2t
5
2−6= 2 t
5
−6 = t
5
t=−30
Thus, the decay constant is −30.
Step 3: Calculate the age of the sample To find the age of the sample, we
use the decay constant in the formula:
t=−
ln N(t)
N0
λ
Substitute the known values:
t=−ln 6
10
−30
t=−ln(0.6)
−30
t≈12.7 days
Therefore, the decay constant is -30 and the age of the sample is approxi-
mately 12.7 days.
Question 17
Question
A radioactive substance decays so that its mass is reduced by 30
14
Solution
Step 1: Let’s denote the mass of the substance after thours by m(t).
Step 2: Since the substance reduces by 30
m(t) = 100 ×1−30
100t
15
.
Step 3: Substitute t= 45 into the equation to find the mass of the substance
after 45 hours:
m(45) = 100 ×1−30
10045
15
.
Step 4: Simplify the expression and calculate the mass:
m(45) = 100 ×1−3
103
= 100 ×7
103
= 100 ×343
1000 = 34.3 grams.
Step 5: Therefore, the mass of the substance after 45 hours is 34.3 grams.
Question 18
Question
A sample of radioactive material initially contains 2.5 grams of a certain isotope.
After 6 hours, only 0.3125 grams of the isotope remain. If the half-life of the
isotope is 3 hours, what is the decay constant λof this isotope? (Hint: Use the
formula N(t) = N0e−λt, where N(t) is the amount of the isotope remaining at
time t,N0is the initial amount, and λis the decay constant.)
Solution
Step 1: We can write the equation for the decay of the isotope as:
N(t) = N0e−λt
where N(t) is the amount of the isotope remaining at time t,N0is the initial
amount, and λis the decay constant.
Step 2: We are given that N0= 2.5 grams, N(6) = 0.3125 grams, and the
half-life T1/2= 3 hours.
Step 3: We can determine the initial decay constant using the half-life for-
mula:
λ=ln(2)
T1/2
Step 4: Substitute the known values into the formula to find λ:
λ=ln(2)
3≈0.2310 hours−1
Step 5: Therefore, the decay constant λof this isotope is approximately
0.2310 hours−1.
15
Question 19
Question
A sample of a radioactive isotope has an initial activity of 500 Bq. After 10
hours, the activity of the sample has decreased to 250 Bq. Calculate the half-life
of the isotope.
Solution
Step 1: Let N0be the initial number of radioactive nuclei, Nbe the number of
radioactive nuclei at time t, and T1/2be the half-life of the isotope. We have
the radioactive decay equation:
N=N0·1
2t
T1/2
Step 2: We are given that the initial activity N0= 500 Bq decreases to
N= 250 Bq after 10 hours. Substituting these values into the equation gives:
250 = 500 ·1
210
T1/2
Step 3: Divide both sides by 500 and take the natural logarithm of both
sides to solve for T1/2:
ln 250
500=10
T1/2
·ln(0.5)
Step 4: Simplify the equation:
−ln(2) = 10
T1/2
·ln(0.5)
Step 5: Divide by ln(0.5) to solve for T1/2:
T1/2=10 ·ln(2)
ln(0.5) ≈10 ·0.693
−0.693 ≈ −10 hours
Step 6: Since the half-life cannot be negative, consider the absolute value:
T1/2≈10 hours
Therefore, the half-life of the radioactive isotope is 10 hours.
Question 20
Question
A certain radioactive substance has a half-life of 30 years. If there are initially
100 grams of the substance, how much will remain after 90 years?
16
Solution
Step 1: The formula to calculate the amount of a radioactive substance remain-
ing after a certain time is given by:
N(t) = N01
2t
T1/2
where: - N(t) is the amount of the substance remaining after time t, - N0is the
initial amount of the substance, - T1/2is the half-life of the substance.
Step 2: Substituting the given values into the formula, we get:
N(90) = 100 1
290
30
Step 3: Simplifying the expression, we have:
N(90) = 100 1
23
= 100 1
8=100
8= 12.5 grams
Therefore, after 90 years, there will be 12.5 grams of the radioactive sub-
stance remaining.
Question 21
Question
A certain radioactive substance has a half-life of 10 days. If you start with 100
grams of the substance, how much will remain after 30 days?
Solution
Step 1: Determine the fraction of the substance remaining after 30 days. The
fraction of the substance remaining after a certain amount of time can be cal-
culated using the formula:
Amount remaining = 1
2time
half-life
In this case:
Amount remaining = 1
230
10
=1
23
=1
8
Step 2: Calculate the amount of the substance remaining after 30 days. If
you start with 100 grams of the substance, the amount remaining after 30 days
can be found by multiplying the initial amount by the fraction remaining:
Amount remaining = 100 ×1
8= 12.5 grams
Therefore, after 30 days, there will be 12.5 grams of the radioactive substance
remaining.
17
Question 22
Question
A certain radioactive material decays according to the function A(t) = A0e−kt,
where A(t) is the amount of material at time t,A0is the initial amount of
material, and kis a positive constant. If 80% of the material remains after 10
years, find the value of k.
Solution
Step 1: Given the information, we can set up the equation A(10) = 0.80A0.
A(10) = A0e−10k= 0.80A0
Step 2: Divide by A0to solve for e−10k.
e−10k= 0.80
Step 3: Take the natural logarithm of both sides.
lne−10k= ln(0.80)
Step 4: Use the properties of logarithms to simplify the left side.
−10k= ln(0.80)
Step 5: Divide by −10 to solve for k.
k=ln(0.80)
−10
Step 6: Calculate the value of kusing a calculator.
k≈ln(0.80)
−10 ≈ −0.0721
Therefore, the value of kis approximately −0.0721.
Question 23
Question
A radioactive substance decays according to the equation N(t) = N0e−0.03t,
where N(t) represents the quantity of the substance at time t(in years), N0is
the initial quantity of the substance, and tis the time elapsed since the initial
quantity was measured. If the initial quantity of the substance is 100 grams,
determine how long it will take for the quantity of the substance to reduce to
50 grams.
18
Solution
Step 1: Substitute the given values into the decay equation to find the time it
takes for the substance to reduce to 50 grams.
50 = 100 ·e−0.03t
50
100 =e−0.03t
0.5 = e−0.03t
Step 2: Take the natural logarithm of both sides to solve for t.
ln(0.5) = lne−0.03t
ln(0.5) = −0.03t
t=ln(0.5)
−0.03
t≈23.1 years
Thus, it will take approximately 23.1 years for the quantity of the substance
to reduce to 50 grams.
Question 24
Question
A sample of a radioactive substance decays over time according to the model
Q(t) = Q0·e−kt, where Q(t) represents the amount of the substance remaining
at time t,Q0is the initial amount of the substance, and kis the decay constant.
Given that the initial amount of the substance is 100 grams and after 10 hours
only 30 grams remain, determine the decay constant k.
Solution
Step 1: Given that Q(t) = Q0·e−kt, we can use the information provided at
t= 0 and t= 10 to set up two equations. At t= 0: Q(0) = 100 = Q0·e0=Q0
At t= 10: Q(10) = 30 = 100 ·e−10k
Step 2: Divide the equation at t= 10 by the equation at t= 0 to eliminate
Q0.30
100 =e−10k
Step 3: Simplify the equation.
0.3 = e−10k
Step 4: Take the natural logarithm of both sides to solve for k.
ln(0.3) = lne−10k=−10k
19
Step 5: Solve for k.
k=ln(0.3)
−10 ≈0.1054
Therefore, the decay constant k≈0.1054.
Question 25
Question
A sample of radioactive material decays over time according to the function
N(t) = N0e−kt, where N(t) represents the amount of material left after tyears,
N0is the initial amount of material, and kis a decay constant.
Given that an initial sample of a radioactive material has a half-life of 10
years, find the amount of material left after 30 years.
Solution
Step 1: Find the decay constant kusing the half-life given.
The half-life of a radioactive material is the time it takes for half of the
material to decay. In this case, we have N(t) = N0
2when t= 10 years.
Substitute N(t) = N0
2and t= 10 years into the decay function:
N0
2=N0e−10k
Solving for k:
1
2=e−10k
ln 1
2= lne−10k
ln 1
2=−10k
k=−1
10 ln 1
2
k≈0.0693 (rounded to 4 decimal places)
Step 2: Find the amount of material left after 30 years using the decay
function.
Substitute N0= 1 (let’s assume 1 unit for easier calculation) and t= 30
years into the decay function:
N(30) = 1 ·e−0.0693·30
Calculating N(30):
N(30) = e−2.079
N(30) ≈0.125 (rounded to 3 decimal places)
20
Therefore, after 30 years, approximately 0.125 units of the radioactive ma-
terial remain.
Question 26
Question
The radioactive isotope iodine-131 decays into xenon-131 with a half-life of 8
days. If a sample initially contains 250 grams of iodine-131, how much will
remain after 32 days?
Solution
Step 1: Determine the decay constant. Step 2: Use the decay constant to find
the amount remaining after 32 days.
Step 1: The decay constant λcan be calculated using the formula:
λ=ln(2)
T1/2
where T1/2is the half-life of the isotope. Substitute T1/2= 8 days into the
formula:
λ=ln(2)
8≈0.0866 days−1
Step 2: The amount remaining after time tcan be calculated using the
formula:
N(t) = N0·e−λt
where: - N(t) is the amount remaining after time t, - N0is the initial amount,
-λis the decay constant, and - tis the time in days. We are given N0= 250
grams, t= 32 days, and λ= 0.0866 days−1. Substitute these values into the
formula:
N(32) = 250 ·e−0.0866·32
N(32) = 250 ·e−2.7712
N(32) ≈250 ·0.0625
N(32) ≈15.625 grams
Therefore, approximately 15.625 grams of iodine-131 will remain after 32
days.
21
Question 27
Question
An unknown radioactive substance decays according to the equation A(t) =
A0e−kt where A(t) is the amount of substance at time t(in years), A0is the
initial amount of substance, and kis the decay constant. Given that the half-life
of the substance is 10 years, find the decay constant k.
Solution
Step 1: Recall that the half-life of a radioactive substance is the time it takes
for half of the initial amount to decay. Using the half-life information given, we
can write the following equation:
A0
2=A0e−k·10
Step 2: Divide both sides by A0to simplify the equation:
1
2=e−10k
Step 3: Take the natural logarithm of both sides of the equation to eliminate
the exponential:
ln 1
2= lne−10k
Step 4: Use the property of logarithms that lnab=bln(a) to simplify the
right side:
ln 1
2=−10kln(e)
Step 5: Recall that ln(e) = 1, so the equation simplifies to:
ln 1
2=−10k
Step 6: Solve for the decay constant kby dividing both sides by −10:
k=−ln 1
2
10 =ln(2)
10
Hence, the decay constant kis k=ln(2)
10 .
Question 28
Question
A sample of a radioactive isotope has an initial activity of 2000 decays per
second. After 3 hours, the activity of the sample has decreased to 500 decays
per second. Calculate the half-life of the isotope.
22
Solution
Step 1: Find the decay constant λusing the formula:
N(t) = N0·e−λt
where N(t) is the activity at time t,N0is the initial activity, and tis the time
elapsed.
Given N0= 2000 decays per second, N(t) = 500 decays per second, and
t= 3 hours, we can rearrange the formula to solve for λ:
500 = 2000 ·e−λ·3·3600
500
2000 =e−10800λ
1
4=e−10800λ
Step 2: Solve for the decay constant λ:
ln 1
4=−10800λ
λ=ln(4)
10800 ≈ −2.53 ×10−4s−1
Step 3: Calculate the half-life T1/2using the formula:
T1/2=ln(2)
λ
Substitute the value of λinto the formula:
T1/2=ln(2)
−2.53 ×10−4
T1/2≈2738.7 seconds ≈45.6 minutes
Therefore, the half-life of the isotope is approximately 45.6 minutes.
Question 29
Question
A sample of a radioactive substance decays according to the equation N(t) =
N0·e−0.02t, where N(t) is the number of atoms remaining after tdays, N0is
the initial number of atoms, and tis the time in days. If the initial number of
atoms is 1012, find the rate of decay after 5 days.
23
Solution
Step 1: Calculate the number of atoms remaining after 5 days.
N(5) = N0·e−0.02·5
= 1012 ·e−0.1
≈1012 ·0.9048
≈9.048 ×1011
Step 2: Find the rate of decay after 5 days. The rate of decay is given by
the derivative of N(t) with respect to t, which is dN
dt =−0.02N0·e−0.02t. Now,
substitute t= 5 and N0= 1012 into the equation:
dN
dt t=5
=−0.02 ·1012 ·e−0.02·5
=−0.02 ·1012 ·e−0.1
≈ −0.02 ·1012 ·0.9048
≈ −1.810 ×1011
Therefore, the rate of decay after 5 days is approximately 1.810×1011 atoms
per day.
Question 30
Question
A sample of radioactive element decays according to the equation N(t) =
N0e−0.01t, where N(t) is the amount of the element remaining after tdays,
N0is the initial amount of the element, and tis the time in days. If the ini-
tial amount of the element is 100 grams, what is the amount of the element
remaining after 20 days?
Solution
Step 1: Substitute the given values into the equation to find the amount of the
element remaining.
N(t) = N0e−0.01t
N(20) = 100e−0.01×20
N(20) = 100e−0.2
Step 2: Calculate the amount of the element remaining after 20 days.
N(20) ≈100 ×0.8187
N(20) ≈81.87 grams
Therefore, the amount of the element remaining after 20 days is approxi-
mately 81.87 grams.
24
Question 31
Question
A sample of a radioactive substance has an initial mass of 500 grams. After 30
days, the mass has decreased to 375 grams. If the half-life of the substance is
10 days, what is the decay constant λof the substance?
Solution
Let’s denote the initial mass of the substance as m0= 500 grams, the final
mass as m30 = 375 grams, and the half-life as T1/2= 10 days. We can use the
formula for radioactive decay to determine the decay constant λ.
Step 1: Calculate the fraction of the substance remaining after 30 days.
The fraction of the substance remaining after time tcan be given by mt
m0=e−λt,
where mtis the mass of the substance after time t. Given that m30 = 375 grams
and m0= 500 grams, we have:
375
500 =e−λ×30
Step 2: Calculate the decay constant λ. From Step 1, we have:
3
4=e−30λ
Taking the natural logarithm of both sides gives:
ln 3
4=−30λ
λ=−ln 3
4
30
Step 3: Calculate the decay constant λ.
λ=−ln 3
4
30 ≈0.0513 days−1
Therefore, the decay constant of the substance is approximately 0.0513
days−1.
Question 32
Question
A certain radioactive substance decays at a rate proportional to the amount
of the substance present. Initially, there are 100 grams of the substance, and
10 grams of the substance remain after 20 days. Determine the half-life of the
substance.
25
Solution
Let A(t) denote the amount of the substance present at time t. Since the
substance decays at a rate proportional to its amount, we have the differential
equation dA
dt =−kA,
where kis the decay constant.
Step 1: Solve the differential equation using separation of variables.
ZdA
A=−kZdt =⇒ln |A|=−kt +C,
where Cis the constant of integration.
Step 2: Apply the initial condition, A(0) = 100, to find the value of C.
ln |100|=−k(0) + C=⇒C= ln(100)
Step 3: Use the value of Cto rewrite the equation as A(t) = e−kt+ln(100) =
100e−kt.
Step 4: Apply the second condition that there are 10 grams of the substance
remaining after 20 days to find the value of k.
A(20) = 100e−20k= 10 =⇒e−20k=1
10 =⇒ −20k= ln 1
10
Step 5: Solve for kand then determine the half-life of the substance.
k=−1
20 ln 1
10=ln(10)
20 ≈0.1155
The half-life of the substance is given by T1/2=ln(2)
k.
T1/2=ln(2)
ln(10)
20
=20 ln(2)
ln(10) ≈6.0151 days
Therefore, the half-life of the substance is approximately 6.0151 days.
Question 33
Question
A certain radioactive substance decays according to the formula N(t) = N0e−0.02t,
where N(t) represents the quantity of the substance after tyears and N0is the
initial quantity. If the initial quantity of the substance is 500 grams, determine
the amount of substance remaining after 20 years.
26
Solution
Step 1: Substitute N0= 500 grams and t= 20 years into the given formula.
N(20) = 500e−0.02×20
Step 2: Simplify the expression inside the exponential function.
N(20) = 500e−0.4
Step 3: Using the property of exponents (e−x=1
ex), rewrite the expression.
N(20) = 500
e0.4
Step 4: Calculate the value of e0.4.
e0.4≈1.4918
Step 5: Substitute the value back and calculate N(20).
N(20) = 500
1.4918 ≈335.16 grams
Therefore, the amount of the substance remaining after 20 years is approxi-
mately 335.16 grams.
Question 34
Question
A sample of a radioactive substance has an initial mass of 10 grams. After 500
years, only 2 grams of the substance remain. If the half-life of the substance is
300 years, determine the decay constant and the equation that models the mass
of the substance as a function of time.
Solution
Step 1: Determine the decay constant. Given that the half-life of the substance
is 300 years, we can use the formula for exponential decay to find the decay
constant, denoted by λ. The formula is given by:
N(t) = N0·e−λt,
where N(t) is the remaining mass of the substance at time t,N0is the initial
mass, and λis the decay constant.
We know that N(300) = N0/2 (half-life) and N(500) = 2 grams. Plugging
these values into the formula, we get two equations:
(N0·e−300λ=N0
2
N0·e−500λ= 2
27
Step 2: Solve for the decay constant. By dividing the second equation by
the first, we can eliminate N0:
e−500λ+300λ=2
N0
2
⇒e−200λ= 4 ⇒ −200λ= ln 4 ⇒λ=−ln 4
200
Step 3: Determine the equation modeling the mass. Using the decay constant
found in Step 2, the equation modeling the remaining mass of the substance as
a function of time tis:
N(t) = 10 ·e−ln 4
200 t
Question 35
Question
A certain radioactive isotope decays at a rate proportional to the amount of the
isotope present. The initial amount of the isotope is 100 grams and after 10
days, only 60 grams are left. Calculate the half-life of this isotope.
Solution
Step 1: Let’s denote the half-life of the isotope by T. We know that the amount
of radioactive material Aat time tis given by the formula:
A(t) = A0·e−kt
where A0is the initial amount of the isotope, kis the decay constant, and t
is time.
Step 2: We are given that the initial amount of the isotope is 100 grams,
A0= 100, and after 10 days, only 60 grams are left, A(10) = 60.
Step 3: Substituting the given values into the formula, we have:
60 = 100 ·e−10k
Step 4: Solve for the decay constant k:
e−10k= 0.6
−10k= ln(0.6)
k=−ln(0.6)
10
Step 5: The half-life Tis related to the decay constant kby the formula:
T=ln(2)
k
28
Step 6: Substitute the value of kinto the formula to find the half-life T:
T=ln(2)
−ln(0.6)
10
Step 7: Simplifying, we get:
T=10 ·ln(2)
ln(0.6) ≈18.19 days
Step 8: Therefore, the half-life of this radioactive isotope is approximately
18.19 days.
29
