Question

If 'f' denotes the ratio of the number of nuclei decayed Nd to the number of nuclei at t=0 N0 then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as: [$\lambda$ is the radioactive decay constant]

• A. $-\lambda e^{-\lambda t}$
• B. $\lambda e^{-\lambda t}$
• C. $\lambda (1-e^{-\lambda t})$
• D. $-\lambda (1-e^{-\lambda t})$

Hint:

Remember the fundamental radioactive decay law: $N(t) = N_0 e^{-\lambda t}$. The rate of decay is $\frac{dN}{dt} = -\lambda N$. The question asks for the rate of change of the fraction of decayed nuclei, not the rate of change of remaining nuclei.

Answer 1

The Correct Option is B

By definition, the fraction 'f' is the ratio of the number of nuclei decayed, $N_d$, to the initial number of nuclei, $N_0$.
$f = \frac{N_d}{N_0}$
The number of decayed nuclei at time t is the initial number minus the number of nuclei remaining, N.
$N_d = N_0 - N$
Substituting this expression for $N_d$ into the equation for f:
$f = \frac{N_0 - N}{N_0} = 1 - \frac{N}{N_0}$
According to the law of radioactive decay, $N = N_0 e^{-\lambda t}$.
Substituting this for N in the equation for f:
$f = 1 - \frac{N_0 e^{-\lambda t}}{N_0} = 1 - e^{-\lambda t}$
The rate of change of 'f' with respect to time is its derivative, $\frac{df}{dt}$.
$\frac{df}{dt} = \frac{d}{dt}(1 - e^{-\lambda t})$
$\frac{df}{dt} = 0 - (-\lambda e^{-\lambda t}) = \lambda e^{-\lambda t}$
Thus, the rate of change of 'f' with respect to time is $\lambda e^{-\lambda t}$.