NEET Physics Nuclear Physics And Radioactivity NEET PYQ-Nuclear Physics

The activity of a radioactive sample is measured as \[{{N}_{0}}\] counts per minute at \[t=0\] and \[{{N}_{0}}/e\] counts per minute at \[t=5\] min. The time (in minute) at which the activity reduces to half its value is           [AIPMT (S) 2010]

A) \[\text{lo}{{\text{g}}_{\text{e}}}\text{2/5}\]

B)      \[\frac{5}{{{\log }_{e}}2}\]

C) \[5{{\log }_{10}}2\]

D)      \[5{{\log }_{e}}2\]

Correct Answer: D

Solution :

Fraction remains after n half lives
\[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{t/T}}\]
	Given \[N=\frac{{{N}_{0}}}{e}\Rightarrow \frac{{{N}_{0}}}{e{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{5/T}}\]
or \[\frac{1}{e}={{\left( \frac{1}{2} \right)}^{5/T}}\]
Taking log on both sides, we get
\[\log 1-\log e=\frac{5}{T}\log \frac{1}{2}\]
\[-1=\frac{5}{T}(-\log 2)\]
\[\Rightarrow \] \[T=5{{\log }_{e}}2\]
Now, let t' be the time after which activity reduces to half \[\left( \frac{1}{2} \right)={{\left( \frac{1}{2} \right)}^{t'/5{{\log }_{e}}2}}\]
\[\Rightarrow \] \[t'=5{{\log }_{e}}2\]