Question

A radioactive substance has density $\rho$, volume $V$ and decay constant $\lambda$. If the molecular weight of the substance is $M$, and Avogadro's number is $N_a$, then the radioactivity of the substance after time ${}^{\prime }{t}^{\prime }$ is:

• A. $\left(\frac{\lambda V\rho N_a}{M}\right)(1-e^{\lambda t})$
• B. $\left(\frac{t{N}_{a}V}{\rho M}\right)e^{\lambda t/2}$
• C. $\frac{\lambda N_a}{V\rho M}{e}^{-\lambda t}$
• D. $\frac{\lambda V\rho N_a{e}^{-\lambda t}}{M}$

Answer 1

Answer: (D)

$\frac{\lambda V\rho N_a{e}^{-\lambda t}}{M}$

Total Mass = volume x density

Hence, $M_t=\rho V.........\left(i\right)$

Number of moles in given mass, $n=\frac{M_t}{M}..........\left(ii\right)$

Note: Assumption is made here that the unit of $M_t$ is gm.

Number of molecules of substance, $N_o=n{N}_a..........\left(iii\right)$

From (i), (ii) and (iii),

$N_o=\frac{\rho V{N}_a}{M}...........\left(iv\right)$

By law of radioactive decay, $N=N_o{e}^{-\lambda t}.......\left(v\right)$

Rate of radioactive decay after time t is given by, $\frac{dN}{dt}=-\lambda N..........\left(vi\right)$

From (iv), (v) and (vi),

$\left|\frac{dN}{dt}\right|=\frac{\lambda \rho V{N}_a{e}^{-\lambda t}}{M}$