After two hours, one- sixteenth of starting amount of a certain radioactive isotope remained undecay

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13.Nuclei

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After two hours, one- sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is

A

$15\, minutes$

B

$30 \,minutes$

C

$45 \,minutes$

D

$1 \,hour$

Solution

(b) $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^{t/T}}$

$ \Rightarrow \left( {\frac{1}{{16}}} \right) = {\left( {\frac{1}{2}} \right)^{2/T}} $

$\Rightarrow {\left( {\frac{1}{2}} \right)^4} = {\left( {\frac{1}{2}} \right)^{2/T}}$

$\Rightarrow T = 0.5\;hour = 30\, minutes.$

Standard 12

Physics

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Similar Questions

Following statements related to radioactivity are given below

$(A)$ Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

$(B)$ The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

$(C)$ Slope of the graph of $\log _{e}$ (no. of undecayed nuclei) $Vs$. time represents the reciprocal of mean life time $(\tau)$.

$(D)$ Product of decay constant ( $\lambda$ ) and half-life time $\left(T_{1 / 2}\right)$ is not constant.

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