Question 18: Atoms & Nuclei

Question

A radioactive sample at any instant has its disintegration rate 5000 disintegrations per min. After 5 min, the rate is 1250 disintegrations per min. Then, the disintegration constant (per min) is

Accepted Answer

According to law of radioactivity, the rate of disintegration is given as
$$R = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$$
Given, $$R = 5000$$
$$ \Rightarrow 5000 = \lambda {N_0}{e^{ - \lambda t}}$$ ...... (i)
After 5 min, $$R = 1250$$
$$	herefore$$ $$1250 = \lambda {N_0}{e^{ - \lambda (t + 5)}}$$ ...... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{5000} \over {1250}} = {{{e^{ - \lambda t}}} \over {{e^{ - \lambda (t + 5)}}}}$$
$$ \Rightarrow 4 = {e^{5\lambda }}$$
Taking log both sides, we get
$${\log _e}4 = {\log _e}({e^{5\lambda }})$$
$$ \Rightarrow lo{g_e}4 = 5\lambda  \Rightarrow \lambda  = {1 \over 5}{\log _e}{2^2}$$
$$ \Rightarrow \lambda  = 0.4{\log _e}2$$

Answer

0.4 loge 2

Answer

0.2 loge 2

Answer

0.1 loge 2

Answer

0.8 loge 2

Atoms & Nuclei

A radioactive sample at any instant has its disintegration rate 5000 disintegrations per min. After 5 min, the rate is 1250 disintegrations per min. Then, the disintegration constant (per min) is

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