Question

For the radioactive material half-life period is $600s$. If initially there are 600 molecules, find the time taken for disintegration of 450 molecules and the rate of disintegration.

Answer 1

After every half time period of a material half of its molecules get disintegrated and half of initial molecules remain in the same form. We also know that the half-life period of a material is $\log 2/\lambda$, where $\lambda$ is decay constant. And the disintegration rate is directly proportional to the number of molecules present in radioactive material.

Complete step by step answer:
We know that, after every half-life period of time half of molecules of radioactive material are disintegrated and half of initial molecules remain in original form. Half-life period of a material is $\log 2/\lambda$, where $\lambda$ is decay constant.
Given, the half-life period is $600s$ and initially there are 600 molecules.
Then, $\lambda = \dfrac{{\log 2}}{{{\text{half - life}}}} = \dfrac{{0.693}}{{600}} = 0.001155$
Time taken to disintegrate 450 molecules is equal to double of half-life. After the disintegration of 450 molecules, 150 molecules remain which is equal to ${(1/4)^{th}}$ molecules present initially.
Then time taken is equal to $600 \times 2 = 1200\sec$.
We know the rate of decomposition is directly proportional to the number of molecules present.
$r\alpha N$, where $r$ is rate or activity of material and $N$ is number of molecules present.
Then, $r = \lambda \times N = 0.001155 \times 150 = 0.173$
And the rate of disintegration of material is ${\text{0}}{\text{.173 disintegration/s}}$.
Hence the correct answer is option C.

Note: Radioactive materials are unstable and decay with a first order disintegration process. Here the rate of decay is directly proportional to the number of molecules present. Due to this, radioactive material takes thousands of years to decay completely. Because with time the number of molecules decreases and rate also decreases due to which theoretically it takes infinite to decay completely.