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A freshly prepared radioactive sample of half-lif 1 hour emits radiations that are 128 times as intense as the permissible safe limit. The minimum time after which this sample can be safely used is

  • a)
    14 hours

  • b)
    7 hours

  • c)
    128 hours

  • d)
    256 hours

Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A freshly prepared radioactive sample of half-lif 1 hour emits radiati...
Given information:
- Half-life of the radioactive sample = 1 hour
- Radiations emitted by the sample are 128 times the permissible safe limit.

To find: Minimum time after which the sample can be safely used.

Solution:
Let the initial intensity of radiations emitted by the sample be I₀.
After one half-life, the intensity becomes I₀/2.
After two half-lives, the intensity becomes (I₀/2)/2 = I₀/4.
Similarly, after three half-lives, the intensity becomes I₀/8, and so on.

The intensity of radiation emitted by the sample is given to be 128 times the permissible safe limit.
Let the permissible safe limit be L.
Then, I₀ = 128L.

To find the minimum time after which the sample can be safely used, we need to find the number of half-lives the sample must undergo to reduce its intensity to the permissible safe limit.

Let the number of half-lives the sample undergoes be n.
Then, (1/2)ⁿ I₀ = L.
Substituting I₀ = 128L, we get:
(1/2)ⁿ (128L) = L
=> (1/2)ⁿ = (1/128)
=> 2ⁿ = 128 = 2^7
=> n = 7.

Therefore, the sample must undergo 7 half-lives to reduce its intensity to the permissible safe limit.
Since the half-life of the sample is 1 hour, the minimum time after which the sample can be safely used is:
7 x 1 hour = 7 hours.

Hence, the correct answer is option (B).
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A freshly prepared radioactive sample of half-lif 1 hour emits radiati...
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A freshly prepared radioactive sample of half-lif 1 hour emits radiations that are 128 times as intense as the permissible safe limit. The minimum time after which this sample can be safely used isa)14 hoursb)7 hoursc)128 hoursd)256 hoursCorrect answer is option 'B'. Can you explain this answer?
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