Question

What is the disintegration constant of a radioactive element, if the number of its atoms diminished by 18% in 24 h?
(a) $2.1\times {{10}^{-3}}/s$
(b) $2.1\times {{10}^{5}}/s$
(c) $2.2\times {{10}^{6}}/s$
(d) $2.2\times {{10}^{-6}}/s$

Answer 1

First, we should know the formula which is to be used here of radioactive elements given as $N={{N}_{0}}{{e}^{-\lambda t}}$ . All the values are already given here i.e. $\dfrac{N}{{{N}_{0}}}$ obtained as $\left( 100-18 \right)$ then we have to convert time t in seconds from hours. After substituting, we will get the value of $\lambda$ which is a disintegration constant.

Formula used:
$N={{N}_{0}}{{e}^{-\lambda t}}$

Complete answer:
Here, we should know that particular radionuclides decay at different rates, so each has its own decay constant, $\lambda$ . The expected decay $\dfrac{-dN}{N}$ is proportional to an increment of time dt. Therefore, we get $\dfrac{-dN}{N}=\lambda dt$ .
The negative sign indicated that N decreases as time increases, as each decay event follows one after another. Thus, the equation we get is
$N={{N}_{0}}{{e}^{-\lambda t}}$ …………………………………(1)

${{N}_{0}}$ is the value of N at time t equals 0 , $\lambda$ is disintegration constant
So, here we have 24 hours. On converting into seconds, we get, $t=24\times 60\times 60s$
So, we have to find a value $\dfrac{N}{{{N}_{0}}}$ which we will get as $\left( 100-18 \right)$ .

So, substituting all the values in equation (1), we get
$\dfrac{N}{{{N}_{0}}}={{e}^{-\lambda t}}$
$\dfrac{82}{100}={{e}^{-\lambda \left( 24\times 60\times 60 \right)}}$
On taking natural log both the sides, we get
$\ln \dfrac{100}{82}=\lambda \times 24\times 60\times 60$

On solving, we get
$0.19845=\lambda \times 24\times 60\times 60$
$\dfrac{0.19845}{24\times 60\times 60}=\lambda$
$\lambda =2.2\times {{10}^{-6}}s$
Thus, the disintegration constant of a radioactive element is $\lambda =2.2\times {{10}^{-6}}s$ .

So, the correct answer is “Option D”.

Note:
The formula of $N={{N}_{0}}{{e}^{-\lambda t}}$ N is the total number of particles, t is time and $\lambda$ is decay constant. To find the value of $\lambda$ and t we have to take natural log on both the sides as we did in solution. But students must take care while taking natural log because if students consider log as base then the answer will change such as, $\log \left( \dfrac{100}{82} \right)=0.0861$, while the value of natural log is $\ln \dfrac{100}{82}=0.19845$. So, students must be careful about these.