JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Modern Physics

A radioactive sample S1 having an activity of

5 μ Ci

has twice the number of nuclei as another sample S2 which has an activity of

10 μ Ci

. The half lives of S1 and S2 can be

(A)20 years and 5 years, respectively✓ Correct
(B)20 years and 10 years, respectively
(C)10 years each
(D)5 years each

Answer: (A)

Solution

Activity is given by

A = λ N = \frac{l n 2}{T} N

, where T is the half-life and N is the number of nuclei.

For S1:

5 μ Ci = \frac{l n 2}{T _{1}} (2 N_{0})

For S2:

10 μ Ci = \frac{l n 2}{T _{2}} (N_{0})

Dividing these equations:

\frac{5}{10} = \frac{T _{2}}{T _{1}} \cdot \frac{2 N _{0}}{N _{0}}

\frac{1}{2} = \frac{2 T _{2}}{T _{1}}

T_{1} = 4 T_{2}

Option (A) satisfies this: 20 years = 4 × 5 years.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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