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The mean lives of radioactive substances are 1620 y and 405 y for α- emission and β- emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α- emission and β- emission simultaneously.
- a)449 y
- b)399 y
- c)549 y
- d)579 y
Correct answer is option 'A'. Can you explain this answer?
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The mean lives of radioactive substances are 1620 y and 405 y for α- ...
Let at some instant of time tt, number of atoms of the radioactive substance are N. It may decay either by α - emission on by β- emission. So, we can write,
If the effective decay constant is λ, then
Now, 
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The mean lives of radioactive substances are 1620 y and 405 y for α- ...
Given data:
Mean life for α-emission (T1) = 1620 years
Mean life for β-emission (T2) = 405 years
Calculation:
Let's assume the initial number of nuclei in the sample is N.
Time taken for 3/4th of the sample to decay by α-emission:
For α-emission, the decay constant (λ1) = 1/T1 = 1/1620
After time t, N/4 nuclei remain undecayed.
Using the formula for radioactive decay: N(t) = N0 * e^(-λ1*t)
N/4 = N * e^(-λ1*t)
1/4 = e^(-λ1*t)
ln(1/4) = -λ1*t
t = ln(4) / λ1
Time taken for 3/4th of the sample to decay by β-emission:
For β-emission, the decay constant (λ2) = 1/T2 = 1/405
After time t, N/4 nuclei remain undecayed.
Using the formula for radioactive decay: N(t) = N0 * e^(-λ2*t)
N/4 = N * e^(-λ2*t)
1/4 = e^(-λ2*t)
ln(1/4) = -λ2*t
t = ln(4) / λ2
Total time for 3/4th of the sample to decay simultaneously:
Since both processes are happening simultaneously, the total time taken for 3/4th of the sample to decay is the maximum of the times calculated above.
t_total = max(t1, t2)
Substitute the values of λ1 and λ2:
t_total = max(ln(4) / (1/1620), ln(4) / (1/405))
t_total = max(449.33, 109.33)
t_total = 449.33 years
Therefore, the time during which three-fourth of the sample will decay if it is decaying both by α-emission and β-emission simultaneously is 449 years, which is option (a).
Mean life for α-emission (T1) = 1620 years
Mean life for β-emission (T2) = 405 years
Calculation:
Let's assume the initial number of nuclei in the sample is N.
Time taken for 3/4th of the sample to decay by α-emission:
For α-emission, the decay constant (λ1) = 1/T1 = 1/1620
After time t, N/4 nuclei remain undecayed.
Using the formula for radioactive decay: N(t) = N0 * e^(-λ1*t)
N/4 = N * e^(-λ1*t)
1/4 = e^(-λ1*t)
ln(1/4) = -λ1*t
t = ln(4) / λ1
Time taken for 3/4th of the sample to decay by β-emission:
For β-emission, the decay constant (λ2) = 1/T2 = 1/405
After time t, N/4 nuclei remain undecayed.
Using the formula for radioactive decay: N(t) = N0 * e^(-λ2*t)
N/4 = N * e^(-λ2*t)
1/4 = e^(-λ2*t)
ln(1/4) = -λ2*t
t = ln(4) / λ2
Total time for 3/4th of the sample to decay simultaneously:
Since both processes are happening simultaneously, the total time taken for 3/4th of the sample to decay is the maximum of the times calculated above.
t_total = max(t1, t2)
Substitute the values of λ1 and λ2:
t_total = max(ln(4) / (1/1620), ln(4) / (1/405))
t_total = max(449.33, 109.33)
t_total = 449.33 years
Therefore, the time during which three-fourth of the sample will decay if it is decaying both by α-emission and β-emission simultaneously is 449 years, which is option (a).
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The mean lives of radioactive substances are 1620 y and 405 y for α- emission and β- emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α- emission and β- emission simultaneously.a)449 yb)399 yc)549 yd)579 yCorrect answer is option 'A'. Can you explain this answer?
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