Question

The half-life of $^{32}P$ is 14.3 days. The specific activity of phosphorus containing specimen having 1.0 part per million $^{32}P$ (At.Wt.of P=31) is:
A) [0.295 Ci/g]
B) [0.595 Ci/g]
C) [0.795 Ci/g]
D) None of these

Answer 1

Hint:. The answer to this question is based on the rate of disintegration of the radioactive element given by $R=\lambda N $ where $\lambda $ is the decay constant and then calculating specific activity in terms of curies.

Complete step by step answer:
In our chapters of chemistry, we have studied about the concepts of radioactivity of an element and their applications in several fields.
Let us now know what is the base of the radioactivity and the rate of disintegration.
- Radioactivity was accidentally discovered by Becquerel which is the process of disintegration of unstable nucleus which undergoes decay. This is the nuclear phenomenon.

- Based on the types of decay that is alpha, beta, gamma decays, law of radioactivity decay was given which states that ‘when a radioactive element undergoes $\alpha ,\beta $ or $\ gamma $ decay, the number of nuclei undergoing the decay per unit time is directly proportional to the total number of nuclei in the sample material.
Thus, if we take N as total number of nuclei in sample and dN is the change in number of nuclei in the sample at time dt then according to the law, we can say rate of change of N or rate of disintegration is,
$R = \lambda N$ ………(1)
where $\lambda $ is the decay constant.
Now, specific activity of a radioactive nucleus is the activity per kg of that material, that is
1kg = 1000g of $^{31}P$ = $\dfrac{1000}{31}{{N}_{0}}$ of $^{31}P$
But the content of $^{32}P$ is one parts per million that is ${{10}^{6}}$ atoms
Therefore, $^{32}P$ = $\dfrac{1000}{31}{{N}_{0}}\times \dfrac{1}{{{10}^{6}}} = \dfrac{1000\times 6.023\times {{10}^{23}}}{31\times {{10}^{6}}}$ [since ${{N}_{0}}$ = Avogadro number]
Thus, N = $1.942\times {{10}^{19}}$ atoms of $^{32}P$
From equation number (1)
\[R=\lambda N\] and $\lambda =\dfrac{0.693}{{{T}_{{}^{1}/{}_{2}}}}$
Thus by substituting values, we have
\[R=\dfrac{0.693\times 1.942\times {{10}^{19}}}{14.3\times 24\times 3600} = 1.089\times {{10}^{13}}{{s}^{-1}}\]

Therefore, specific activity = $\dfrac{1.089\times {{10}^{13}}}{3.7\times {{10}^{10}}}$ = 294.32 Ci/kg [since 1 curie = $3.7\times {{10}^{10}}$ decay/sec]
In terms of grams, the answer is $\approx $ 0.29Ci/g
So, the correct answer is “Option A”.

Note: Note that the rate of disintegration of radioactive nuclei is also given in the form $-\dfrac{dN}{dt}=\lambda N$ the L.H.S. quantity is written as ‘R’ and not to be confused with this term and also 1 curie and 1 becquerels are one and the same. Becquerel replaced curie as a radiation unit as per International SI.