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

Hint: 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:
\[A = \lambda N,\,N = \dfrac{A}{\lambda }\]
\[1Ci = 3.7 \times {10^{10}}dps\]
\[A = 3.7 \times {10^{10}} \times 3 \times {10^{ - 6}}\]
\[N = 3.7 \times {10^4} \times 3 = 11.1 \times {10^4}\]