Question

If the activity $^{108}Ag$ is 3 microcurie, the number of atoms present in it are $\left( {\lambda = 0.005{{\sec }^{ - 1}}} \right)$
A. $22 \times {10^7}$
B. $2.2 \times {10^6}$
C. $2.2 \times {10^5}$
D. $2.2 \times {10^4}$

Answer 1

The activity of a sample is the average number of disintegrations per second its unit is the Becquerel (Bq). One Becquerel is one decay per second. The decay constant 1 is the probability that a nucleus will decay per second, so its unit is ${s^{ - 1}}$. The half life is the time for half the nuclei to decay.

Complete step by step answer:
Given:
$\lambda = 0.005{\sec ^{ - 1}}$
1 Microcurie $= 3.7 \times {10^4}\,dps$
Assuming all Ag atoms to be radioactive, and each showing.
Let the number of atoms present = 1 dps no.
Given:
$\left( {\dfrac{{dN}}{{dt}}} \right)\;activity = \lambda {N_0} \Rightarrow \,3.7 \times 3 \times {10^4}dps = 0.005 \times {N_0}.$
$\Rightarrow \,{N_0} = 2.2 \times {10^7}\;atoms.$

So, the correct answer is “Option A”.

Note: We also solve as:
$A = \lambda N,\,N = \dfrac{A}{\lambda }$
$1Ci = 3.7 \times {10^{10}}dps$
$A = 3.7 \times {10^{10}} \times 3 \times {10^{ - 6}}$
$N = 3.7 \times {10^4} \times 3 = 11.1 \times {10^4}$