Question

The radioactive decay constant of $_{38}^{90}Sr$ is $7 \cdot 8 \times {10^{ - 10}}{s^{ - 1}}$. The activity of 15 gm of this isotope will be.
A. $1 \cdot 5Ci$
B. $2 \cdot 13 \times {10^3}Ci$
C. $7 \cdot 88Ci$
D. $8 \cdot 76Ci$

Answer 1

Hint:Before we understand the meaning of the term activity, it is important to understand the phenomenon of radioactivity. The radioactivity is the phenomenon by which large unstable nuclei of an element undergo dissociation to produce energy through radiation.

Step by step solution:
The radioactive materials have unstable nuclei. Hence, over time, they dissociate to form two smaller nuclei which releases energy in the form of particles such as alpha or beta particles or even gamma radiation, which results due to emission from nucleus at a higher energy state.
The phenomenon of radioactivity is random and hence, we have to look at it statistically.
If there are ${N_0}$ numbers of nuclei in a material at a particular unit of time, the numbers decrease to a number $N$ after the time t. The relationship among these numbers is given by –
$N = {N_0}{e^{ - \lambda t}}$
where,
$\lambda $ is called the decay constant, which is equal to the time taken for a number of particles to be reduced to 37% of itself, by radioactive decay.
If we differentiate the above equation with respect to time, we get –
$\dfrac{{dN}}{{dt}} = - \lambda \left( {{N_0}{e^{ - \lambda t}}} \right)$
$\dfrac{{dN}}{{dt}} = - \lambda N$
This quantity $\dfrac{{dN}}{{dt}}$ is called as activity which represents the number of disintegrations of the nuclei per unit time.
Activity, $A = \left| {\dfrac{{dN}}{{dt}}} \right| = \lambda N$
The SI unit of activity is becquerel (Bq).


Given,
Decay constant of strontium $_{38}^{90}Sr$, $\lambda = 7 \cdot 8 \times {10^{ - 10}}{s^{ - 1}}$
Mass of strontium, $m = 15mg$
To find the number of atoms of strontium in 15 mg of strontium, we have to divide the given mass by the molar mass of strontium.
Molar mass of strontium, $M = 90g$
Number of moles of strontium in 15mg, $n = \dfrac{m}{M} = \dfrac{{15}}{{90}} = \dfrac{1}{6}$
One mole of a substance contains $6 \cdot 023 \times {10^{23}}$ atoms.
Hence, number of atoms in the given substance,
$N = 6 \cdot 023 \times {10^{23}} \times \dfrac{1}{6} = 1 \cdot 004 \times {10^{23}}$
Therefore, the activity is given by,
$A = \lambda N = 7 \cdot 8 \times {10^{ - 10}} \times 0 \cdot 27 \times {10^{23}} = 7 \cdot 8312 \times {10^{13}}Bq$
There is another older unit of radioactivity which was in use before becquerel, called curie. The relationship between these units is given by –
$1Ci = 3 \cdot 7 \times {10^{10}}Bq$
The activity, in the units of curie,
$A = \dfrac{{7 \cdot 8312 \times {{10}^{13}}}}{{3 \cdot 7 \times {{10}^{10}}}} = 2 \cdot 13 \times {10^3}Ci$

Hence, the correct option is Option B.

Note:The unit of activity curie, is named after the Polish scientist Marie Curie, who is known for her pioneering work in the field of radioactivity, on whose honour, the unit was named. But, later, the SI unit of activity was named becquerel after the French scientist Henry Becquerel who is famous for his discovery of radioactivity in uranium.