CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The activity for a sample of radioactive element found in an experiment is $$6400$$ decays per minute. Repeating this experiment after $$6$$ days, activity was found to be $$400$$ decays per minute. Calculate half life of the given element.


Solution

We know that 
Activity $$R=\lambda N$$
where,
$$N=N {_0e^{-\lambda t}}$$
$$R=\lambda N {_0e^{-\lambda t}}$$
$$R=R {_0e^{-\lambda t}}$$
$$\dfrac{R}{R_0}=e^{-\lambda t}$$....................$$(1)$$
From Rutherford soddy law, $$N=N {_0e^{-\lambda t}}$$
$$\dfrac{N}{N_0}= e^{-\lambda t}$$......$$(2)$$
From equation $$(1)$$ and $$(2)$$,
$$\dfrac{R}{R_0}=\dfrac{N}{N_0}$$
where, $$N=\dfrac{N_0}{2^n}$$        $$\Rightarrow \dfrac{R}{R_0}=\dfrac{1}{2^{t/T_{1/2}}}$$
$$N=\dfrac{N_0}{2^{t/T_{1/2}}}$$
$$\Rightarrow \dfrac{1}{16}=\dfrac{1}{2^{6/T_{1/2}}}$$
$$\dfrac{N}{N_0}=\dfrac{1}{2^{t/T_{1/2}}}$$
$$\Rightarrow \dfrac{1}{2^4}=\dfrac{1}{2^{6/T_{1/2}}}$$
Given,
$$\dfrac{R}{R_0}=\dfrac{400}{6400}$$     $$\Rightarrow \dfrac{R}{R_0}=\dfrac{1}{16}$$
$$t=6days$$
Comparing powers
$$\dfrac{6}{T_{1/2}}=4$$
$$T_{1/2}=\dfrac{6}{4}$$
$$T_{1/2}=1.5days$$

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image