Question

A sample of radioactive material decays simultaneously by two processes A and B with half lives $\frac{1}{2}$ and $\frac{1}{4}$ hr respectively. For first half hr it decays with the process A, next one hr with the process B and for further half an hour with both A and B. If originally there were N₀ nuclei, find the number of nuclei after 2 hr of such decay.

• A. $\frac{N_{0}}{(2)^{8}}$
• B. $\frac{N_{0}}{(2)^{4}}$
• C. $\frac{N_{0}}{(2)^{6}}$
• D. $\frac{N_{0}}{(2)^{5}}$

Answer 1

Answer: (A)

$\frac{N_{0}}{(2)^{8}}$

Answer 2

After first half hrs N = N₀ . $\frac { 1 } { 2 }$

for t = $\frac { 1 } { 2 }$hr to t = 1 $\frac { 1 } { 2 }$ hrs

N = $\left[ \mathrm { N } _ { 0 } \cdot \frac { 1 } { 2 } \right]$ $\left[ \frac { 1 } { 2 } \right] ^ { 4 }$ = N₀ $\left[ \frac { 1 } { 2 } \right] ^ { 5 }$

For t = 1$\frac { 1 } { 2 }$ hrs to t = 2 hrs. [for both A and B

$\frac { 1 } { t _ { 1 / 2 } } = \frac { 1 } { 1 / 2 } + \frac { 1 } { 1 / 4 }$= 2 + 4 = 6 Ž t_1/2 = 1/6 hrs.]

N = $\left[ \mathrm { N } _ { 0 } \left( \frac { 1 } { 2 } \right) ^ { 5 } \right] \left[ \frac { 1 } { 2 } \right] ^ { 3 } = \mathrm { N } _ { 0 } \left[ \frac { 1 } { 2 } \right] ^ { 8 }$