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Question: In a mean life of a radioactive sample :- (A) About 1/3 of substance disintegrate (B) Ab...

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

Explanation

Solution

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

(N)=Noeλt\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(N)=Noeλt\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 (N)=Noeλ1λ\left( N \right) = {N_o}{e^{ - \lambda \dfrac{1}{\lambda }}}

Here will cancel so we get

N=Noe1N = {N_o}{e^{ - 1}}
Or we can write N=NoeN = \dfrac{{{N_o}}}{e}..........................(1)
Now we know that NNoN - {N_o}
So in equation (1)

N=No3N = \dfrac{{{N_o}}}{3}

Substance disintegration will be difference between original value(N) to obtained value(No)
ie: 2No3\dfrac{{2{N_o}}}{3}
substituting the value of N in above equation

NoNo3{N_o} - \dfrac{{{N_o}}}{3}

Therefore substance disintegrated is 2No3\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.