Question

A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular second. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is
A. $0.2\log _e^2$
B. $0.5\log _e^2$
C. $0.6\log _e^2$
D. $0.5{\text{l}}o{g_e}2$

Answer 1

Use the equations for the rate of disintegrations of the radioactive element and the decay rate equation. These equations give the relation between decay constant, decay rate and time.

Formulae used:
The expression for the rate of disintegrations of the radioactive element is
$\dfrac{{dN}}{{dt}} = - \lambda N$ …… (1)
Here, $\dfrac{{dN}}{{dt}}$ is the rate of disintegration of the radioactive element, $\lambda$ is the decay constant and $N$ is the population of the radioactive element at time $t$.
The negative sign indicates that the population of the radioactive element is decreasing.
The decay rate equation is
$N = {N_0}{e^{ - t\lambda }}$ …… (2)
Here, $N$ is the population of the radioactive element at time $t$and is the initial population of the radioactive element before disintegration.

Complete step by step answer:
Rewrite the equation for the rate of disintegrations of the radioactive element for the initial rate 10,000 disintegrations per minute for ignoring the negative sign.
Substitute $10000$ for $\dfrac{{dN}}{{dt}}$ in equation (1).
$\Rightarrow 10000 = \lambda {N_0}$
Here, ${N_0}$ is the initial population of the radioactive element.
Rewrite the equation for the rate of disintegrations of the radioactive element for the rate 2500 disintegrations per minute for the fourth minute ignoring the negative sign.
Substitute $2500$ for $\dfrac{{dN}}{{dt}}$ in equation (1).
$\Rightarrow 2500 = \lambda N$
Here, $N$ is the population of the radioactive element at the fourth minute.
Rewrite equation (2) for the fourth minute.
Substitute $4\,{\text{min}}$ for $t$ in equation (2).
$\Rightarrow N = {N_0}{e^{ - \left( {4\,{\text{min}}} \right)\lambda }}$
Multiply on both sides of the above equation by the decay constant $\lambda$ .
$\Rightarrow N\lambda = {N_0}\lambda {e^{ - \left( {4\,{\text{min}}} \right)\lambda }}$
Substitute for $N\lambda$ and for ${N_0}\lambda$ in the above equation.
$\Rightarrow 2500 = 10000{e^{ - \left( {4\,{\text{min}}} \right)\lambda }}$
$\Rightarrow \dfrac{1}{4} = {e^{ - 4\lambda }}$
Take natural log on both sides of the above equation.
$\Rightarrow {\log _e}\dfrac{1}{4} = - 4\lambda$
$\Rightarrow {\log _e}1 - {\log _e}4 = - 4\lambda$
$\Rightarrow 0 - {\log _e}4 = - 4\lambda$
$\Rightarrow {\log _e}4 = 4\lambda$
$\Rightarrow {\log _e}{2^2} = 4\lambda$
$\Rightarrow 2{\log _e}2 = 4\lambda$
$\Rightarrow \lambda = \dfrac{1}{2}{\log _e}2$
$\therefore \lambda = 0.5{\text{l}}o{g_e}2$

Hence, the decay constant per minute is $0.5{\text{l}}o{g_e}2$.Hence, the correct option is D.

Note: The rate of disintegrations 10000 disintegrations per minute is considered the initial rate of disintegration. For a particular radioactive element, the decay constant remains the same. Hence, the decay constant for the given reaction is kept the same for all times.