Question

The half life of $^{238}U$ for $\alpha - $ decay is $4.5 \times {10^9}$ years. The number of disintegration per second occur in 1g of $^{238}U$ is
(Avogadro’s number = $6.023 \times {10^{23}}mo{l^{ - 1}}$)
(A) $1.532 \times {10^4}{s^{ - 1}}$
(B) $1.325 \times {10^4}{s^{ - 1}}$
(C) $1.412 \times {10^4}{s^{ - 1}}$
(D) $1.235 \times {10^4}{s^{ - 1}}$

Answer 1

Hint: In 238 gm of uranium Avogadro numbers of atoms are present. So we need to find the number of atoms contained in 1gm of uranium. Then from the half-life given we find the decay rate of uranium and from there, the number of disintegrations per second will be the product of the number of atoms and the decay rate.
Formula used: We will be using the following formulas,
$\Rightarrow \lambda = \dfrac{{\ln 2}}{{{t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}}}}$
where $\lambda $ is the decay rate and ${t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}}$ is the half-life of the species.

Complete step by step answer:
We are given 1gm of $^{238}U$. Now from the mole concept, we can say that 1 mole of uranium contains 238 gm of uranium. And again 1 mole of uranium has Avogadro number of atoms. Therefore,
238 gm of $^{238}U$ = $6.023 \times {10^{23}}$ atoms.
So the number of atoms contained in 1 gm of uranium is found by dividing both the sides of the equation by 238,
1gm of $^{238}U$ = $\dfrac{{6.023 \times {{10}^{23}}}}{{238}}$ atoms.
this on calculation gives
$2.530 \times {10^{21}}$ atoms.
Now the half-life of the uranium is given as, ${t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}} = 4.5 \times {10^9}$ years.
To convert this value into seconds we multiply $3.16 \times {10^7}$.
Hence we get,
$\Rightarrow {t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}} = 4.5 \times {10^9} \times 3.16 \times {10^7}$
On doing the product we get,
$\Rightarrow {t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}} = 1.422 \times {10^{17}}$ seconds.
Therefore the decay rate of uranium can be found from the expression,
$\Rightarrow \lambda = \dfrac{{\ln 2}}{{{t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}}}}$
substituting the value of the half-life, we get,
$\Rightarrow \lambda = \dfrac{{\ln 2}}{{1.422 \times {{10}^{17}}}}$
On doing the above calculation the value of the decay rate is obtained as,
$\Rightarrow \lambda = 4.87 \times {10^{ - 18}}{s^{ - 1}}$
So now, the number of disintegrations per second is given by the product of the decay rate and the number of atoms in 1gm of uranium.
Number of disintegrations = $4.87 \times {10^{ - 18}} \times 2.530 \times {10^{21}}$
By doing the product we get,
The number of disintegrations = $1.235 \times {10^4}$ atoms per second.
Hence the correct option will be (D); $1.235 \times {10^4}{s^{ - 1}}$.

Note:
The half-life of uranium is the time required for half the population of uranium atoms to decay and for any radioactive sample the half-life and the decay constant are related by the formula, $\lambda = \dfrac{{\ln 2}}{{{t_{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}}}}$. With the help of the decay rate, we can use the decay equation to find the concentration of atoms in a radioactive species after a certain time.