Question

In a nuclear reactor the activity of a radioactive substance is $2000 / s$. If the mean life of the products is $50$ minutes, then in the steady power generation, the number of radio nuclides is

• A. $12 \times 10^5$
• B. $60 \times 10^5$
• C. $90 \times 10^5$
• D. $15 \times 10^5$

Answer 1

Answer: (B)

$60 \times 10^5$

Answer 2

Given,
nuclcar rcactor the activity of a radioactivc substance,
$\frac{d N}{d t}=2000 / s$
and mean-life of the products,
$\tau =50\, min$
$=50 \times 60 \,sec$
Now, the mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$ i.e.,
$\tau=\frac{1}{\lambda}$
$\Rightarrow \lambda=\frac{1}{\tau}=\frac{1}{50 \times 60}$ per second
$\therefore$ The rate of decay is proportional to the number of radio nuclides is given as
$\left|\frac{d N}{d t}\right| \propto N$
$\frac{d N}{d t} \mid=\lambda\, N$
$\Rightarrow 2000=\frac{1}{50 \times 60} \times N$
Where, $\lambda$ is a disintegration constant.
Putting the given values, we get
$N=2000 \times 50 \times 60$
$\Rightarrow N=60 \times 10^{5}$
Hence, the number of nuclides is $60 \times 10^{5}$