Question

Specific activity of a radioisotope is given by :
(A) $\dfrac{\lambda \text{ x }N}{W}$
(B) $\dfrac{\lambda \text{ x }{{N}_{A}}}{M}$
(C) $\dfrac{4.17\text{ x 1}{{\text{0}}^{23}}}{M\text{ x }{{t}_{1/2}}}$
(D) none of the above

Answer 1

Hint: Specific activity is a physical property of the radioactive isotope. It is a constant value even if the value of activity changes. This is because the specific activity is the activity of the radioisotope per unit mass. Write the general formula for activity and then divide by the molecular mass to obtain specific activity.

Complete step-by-step answer:
Activity is the number of disintegrations per second, or rather the number of unstable atomic nuclei that decay per second for a given sample.
 Activity is determined by radiation detectors and electronic circuits. The number of particles and photons i.e. pulses of electromagnetic energy ejected from a radioactive material during a convenient time interval.
This experimental count, however, must be interpreted in the light of a thorough knowledge of the particular manner of radioactive decay in the sample material, because some sources emit more than one particle or photon per disintegration which interfere with our calculation of activity of the radioactive sample.
The SI unit for activity is becquerel. One becquerel is defined as the activity of a sample of radioactive material in which one nucleus decays in a second.
Specific activity is the activity of a sample of radioactive material per unit mass.

The formula for activity is given by:
A = $\lambda \text{ x }N$
Specific activity = $\dfrac{\lambda \text{ x }N}{W}$
Where,
$\lambda $ is the decay constant,
N is the number of atoms present in the sample,
W is the weight of the total sample.

If we take the number of atoms as 1 mole, W will become equal to the molar mass (M).
Specific activity becomes $\dfrac{\lambda \text{ x }{{N}_{A}}}{M}$
When we substitute the values of Avogadro number and $\lambda $ as $\dfrac{\ln 2}{{{T}_{1/2}}}$, we get:
Specific activity = $\dfrac{4.17\text{ x 1}{{\text{0}}^{23}}}{M\text{ x }{{t}_{1/2}}}$

Therefore, the correct answers are options are (A), (B) and (C).

Note: Radioactive decay as a controlled reaction is used for the production of electricity by disintegrating uranium atoms. However nuclear reactions can become uncontrolled if proper care is not taken in reducing the decay rate. Uncontrolled nuclear reactions emit radiation that can cause havoc to mankind as the radiation affects the proper functioning of the human body.