Question

A radioactive isotope $_z{X^A}$ is converted into $_{z - 8}{Y^{A - 16}}$ by a-decay. If $1$ gram-atom $_z{X^A}$, produced $3$ mol $2H{e^ + }$in $34.65$ hour then disintegration constant of $_z{X^A}$ is (in $h{r^{ - 1}}$ is ):
(a) $8 \times {10^2}$
(b) $4 \times {10^{ - 2}}$
(c) $2 \times {10^{ - 2}}$
(d) $3.6 \times {10^{ - 3}}$

Answer 1

Hint:For this question, first we have to understand the question and find out what the question is asking, what the question wants to convey by saying three moles are produced, decrease in the atomic number of the element will tell us how many helium are produced.
Formula used:
For first order reaction,
The relation of rate of reaction to the time and initial concentration and final concentration are as follows:
$k = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{A}} \right]}}{t}$
Where, $k$ is rate of reaction or disintegration constant, $t$ is time, ${[A]_o}$ is initial concentration and $A$ is final concentration left.

Complete step by step answer:
Radioactive isotope: these are the isotopes of some radioactive material. The combination of neutrons and protons and a combination of excess energy which is unstable.
Now, what is the definition of disintegration constant,
Disintegration Constant: The disintegration constant is proportionality between the size of the total radioactive atoms and the rate at which the radioactive atoms decrease because of radioactive decay.
The reaction in the given question,
$_z{X^A}{ \to _{z - 8}}{Y^{A - 16}} + {4^4}_2He$
$1$ mole gives $4$ moles of helium,
In the question given, $3$ moles of helium produced.
Hence, the reaction is $\dfrac{3}{4}$ completed, i.e, $\dfrac{1}{4}$ reaction is left.
 Final concentration left= Initial concentration – the reaction completed
$A = {[A]_O} - 0.75{[A]_O} = 0.25{[A]_O}$
So, the final concentration will be $A = 0.25{[A]_O}$ .
Now, putting the value of $t$ and $A$ in the rate equation ( $k = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{A}} \right]}}{t}$ )
$k = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{{0.25{{[A]}_o}}}} \right]}}{t} = \dfrac{{\ln [4]}}{{34.65}}$
 From the properties of $\ln $ , $\ln (4) = 2\ln (2)$ and $\ln (2) = 0.693$ ,
$ \Rightarrow k = \dfrac{{2 \times \ln [2]}}{{34.65}} = \dfrac{{2 \times 0.693}}{{34.65}}$
Now, solving the equation,
We get,
$ \Rightarrow k = 0.04h{r^{ - 1}} = 4 \times {10^{ - 2}}h{r^{ - 1}}$
As the rate of reaction and the disintegration rate are the same.
Hence, the disintegration constant is $4 \times {10^{ - 2}}h{r^{ - 1}}$ .

Hence, the correct option is (b) $4 \times {10^{ - 2}}$ .

Note:
 When the time of any reaction or the time at which that amount of the reaction is completed is given in the question, think about the first order formula and try to find out the disintegration constant by using the half-life formula and be careful what is given the percentage of reaction completed or left.