Question

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.

Answer 1

Correct Answer: 15

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.

1. Set N(t) = 1, N₀ = 64, thus:

1 = 64 × (1/2)^(t/2.5)

 

1. Rearrange the equation to isolate the exponential term:

(1/2)^(t/2.5) = 1/64

 

1. Recognize that 1/64 = (1/2)⁶, hence:

(1/2)^(t/2.5) = (1/2)⁶

 

1. Equate the exponents as the bases are identical:

t/2.5 = 6

 

1. 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.

Answer 2

Given here, T=150 minutes, A₀=64x, where, x is the safe limit.
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