Question

In a mean life of a radioactive sample :-

 (A) About 1/3 of substance disintegrate

(B) About 2/3 of substance disintegrate

(C) About 90% of the substance disintegrate

 (D) Almost all the substance disintegrates

Answer 1

Hint-here we can see that mean life is related with disintegration in all the option so we need to use formula for decay

Formula used

\[\left( N \right) = {N_o}{e^{ - \lambda t}}\]

 Mean life(t)=T=\[N = \dfrac{{{N_o}}}{3}\]\[\dfrac{1}{\lambda }\]

Where is constant of proportionality or decay constant

Complete step by step solution

 We know that expression for decay in terms of mean life

Radioactive nuclei\[\left( N \right) = {N_o}{e^{ - \lambda t}}\]

And Mean life(t)=T=\[\dfrac{1}{\lambda }\]\[\dfrac{1}{\lambda }\]
So putting the value of mean life in the decay formula

Radioactive nuclei \[\left( N \right) = {N_o}{e^{ - \lambda \dfrac{1}{\lambda }}}\]

Here will cancel so we get

\[N = {N_o}{e^{ - 1}}\]
Or we can write \[N = \dfrac{{{N_o}}}{e}\]..........................(1)
Now we know that \[N - {N_o}\]
So in equation (1)

\[N = \dfrac{{{N_o}}}{3}\]

Substance disintegration will be difference between original value(N) to obtained value(No)
ie: \[\dfrac{{2{N_o}}}{3}\]
substituting the value of N in above equation

\[{N_o} - \dfrac{{{N_o}}}{3}\]

Therefore substance disintegrated is \[\dfrac{{2{N_o}}}{3}\]

So option (B) is correct

Note-radioactivity decay is a process which depends upon instability of the particular radioisotope. The decay process and observed half life dependence of radioactivity can be predicted by assuming that individual nuclear decay is a purely random event.