A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _______ hours.
Correct Answer: 15
Approach Solution - 1
To determine the time after which the radiation emitted by the radioactive source falls to a safe level, we begin by understanding radioactive decay characterized by half-life. Given:
- Initial intensity = 64 times the safe level.
- Half-life (T₁/₂) = 2.5 hours (2 hours 30 minutes).
The formula for radioactive decay is:
N(t) = N₀ × (1/2)^(t/T₁/₂)
where N(t) is the intensity at time t, N₀ is the initial intensity. We set N(t) equal to 1 (safe level) and solve for t.
- Set N(t) = 1, N₀ = 64, thus:
1 = 64 × (1/2)^(t/2.5)
- Rearrange the equation to isolate the exponential term:
(1/2)^(t/2.5) = 1/64
- Recognize that 1/64 = (1/2)⁶, hence:
(1/2)^(t/2.5) = (1/2)⁶
- Equate the exponents as the bases are identical:
t/2.5 = 6
- Solve for t:
t = 6 × 2.5 = 15 hours
Thus, the minimum required time is 15 hours. This value lies within the expected range of 15 to 15 hours.
Approach Solution -2
Using the radioactive decay law relation,
A=A0×2\(^{\frac{-t}{T}}\), where, t is the minimum time in hours after which it would be possible to work safely with source.
Putting the values, we have x=64x×2\(^{\frac{-t}{T}}\)
∴ t=6T=6×2.5=15 hours
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Concepts Used:
Half-life
The amount of time taken for half of a particular sample to react is known as Half-life.
Half-Life Formula:
We can describe exponential decay by any of the three formulas
