Question

The half-life of $U^{238}$ for $\alpha$-decay is $4.5\times {10}^9\text{years}$. The number of disintegrations per second that occur in $1\text{g}$ of $U^{238}$ is :
(Avogadro number $=6.023\times {10}^{23}{\text{mo1}}^{-1}$)

• A. $1.725\times {10}^4{\text{s}}^{-1}$
• B. $1.535\times {10}^4{\text{s}}^{-1}$
• C. $1.035\times {10}^4{\text{s}}^{-1}$
• D. $1.235\times {10}^4{\text{s}}^{-1}$

Answer 1

The correct option is D $1.235\times {10}^4{\text{s}}^{-1}$

Given:

$T_{1/2}=4.5\times {10}^9\text{years}=4.5\times 365\times 86400\times {10}^9\text{sec}$

$N_A=6.023\times {10}^{23}{\text{mol}}^{-1}$

$A=238$

$238\text{g of uranium}\to 6.023\times {10}^{23}\text{atoms}$

$1\text{gm of uranium}\to N=\frac{6.023\times {10}^{23}}{238}\text{atoms}$

We know, $T_{1/2}=\frac{0.693}{\lambda }\Rightarrow \lambda =\frac{0.693}{T_{1/2}}$

$\therefore \frac{dN}{dt}=-\lambda N=\frac{0.693}{T_{1/2}}\times N$

$=\frac{0.693\times 6.023\times {10}^{23}}{4.5\times 365\times 86400\times {10}^9\times 238}$

$=\frac{4.174\times {10}^{23}}{3.378\times {10}^{19}}$

$\Rightarrow \frac{dN}{dt}=1.235\times {10}^4{\text{s}}^{-1}$

Hence, $\left(\text{D}\right)$ is the correct answer.

Why this question? Gives an idea to use the relation between decay law and half life period to solve the numerical.