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Radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains?
- a)1080
- b)2430
- c)3240
- d)4860
Correct answer is option 'A'. Can you explain this answer?
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Radioactive material decays by simultaneous emission of two particles ...
Solution:
Given, half-life of first particle = 1620 years
Half-life of second particle = 810 years
To find: Time after which one-fourth of the material remains
Let us assume that initially, we have 1 unit of radioactive material.
After 1 half-life of the first particle, the amount of material remaining = 1/2
After 2 half-lives of the first particle, the amount of material remaining = (1/2) × (1/2) = 1/4
After 3 half-lives of the first particle, the amount of material remaining = (1/2) × (1/2) × (1/2) = 1/8
Similarly, after n half-lives of the first particle, the amount of material remaining = (1/2)^n
Similarly, after 1 half-life of the second particle, the amount of material remaining = 1/2
After 2 half-lives of the second particle, the amount of material remaining = (1/2) × (1/2) = 1/4
Similarly, after n half-lives of the second particle, the amount of material remaining = (1/2)^n
Now, let us assume that after t years, one-fourth of the material remains.
So, the amount of material remaining = (1/4) × 1 = 1/4
The amount of material remaining after t years can be expressed as:
(1/2)^n × (1/2)^m = (1/2)^n+m
where n is the number of half-lives of the first particle and m is the number of half-lives of the second particle.
We need to find n + m, which can be expressed as:
t/1620 + t/810 = t(1/1620 + 1/810) = t(3/3240 + 6/3240) = t/540
(Here, we have used the formula for half-life: t1/2 = (ln2)/λ, where λ is the decay constant)
So, we have:
(1/2)^(t/1620 + t/810) = (1/2)^(t/540) = 1/4
Taking log on both sides, we get:
t/540 = log(1/4)/log(1/2)
t/540 = 2
t = 1080 years
Therefore, the time after which one-fourth of the material remains is 1080 years.
Hence, the correct option is (a) 1080.
Given, half-life of first particle = 1620 years
Half-life of second particle = 810 years
To find: Time after which one-fourth of the material remains
Let us assume that initially, we have 1 unit of radioactive material.
After 1 half-life of the first particle, the amount of material remaining = 1/2
After 2 half-lives of the first particle, the amount of material remaining = (1/2) × (1/2) = 1/4
After 3 half-lives of the first particle, the amount of material remaining = (1/2) × (1/2) × (1/2) = 1/8
Similarly, after n half-lives of the first particle, the amount of material remaining = (1/2)^n
Similarly, after 1 half-life of the second particle, the amount of material remaining = 1/2
After 2 half-lives of the second particle, the amount of material remaining = (1/2) × (1/2) = 1/4
Similarly, after n half-lives of the second particle, the amount of material remaining = (1/2)^n
Now, let us assume that after t years, one-fourth of the material remains.
So, the amount of material remaining = (1/4) × 1 = 1/4
The amount of material remaining after t years can be expressed as:
(1/2)^n × (1/2)^m = (1/2)^n+m
where n is the number of half-lives of the first particle and m is the number of half-lives of the second particle.
We need to find n + m, which can be expressed as:
t/1620 + t/810 = t(1/1620 + 1/810) = t(3/3240 + 6/3240) = t/540
(Here, we have used the formula for half-life: t1/2 = (ln2)/λ, where λ is the decay constant)
So, we have:
(1/2)^(t/1620 + t/810) = (1/2)^(t/540) = 1/4
Taking log on both sides, we get:
t/540 = log(1/4)/log(1/2)
t/540 = 2
t = 1080 years
Therefore, the time after which one-fourth of the material remains is 1080 years.
Hence, the correct option is (a) 1080.
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Community Answer
Radioactive material decays by simultaneous emission of two particles ...
N = N0e-(λ1 + λ2) t.
4 = e(λ1 + λ2) t.
t =
= 1080 year.
4 = e(λ1 + λ2) t.
t =
= 1080 year.
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Radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains?a)1080b)2430c)3240d)4860Correct answer is option 'A'. Can you explain this answer?
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