If f denotes the ratio of the number of nuclei decayed ( N _ d ) to the number of nuclei at t = 0, ( N _ 0 ) then for a collection of radioactive nuclei, the rate of change of f with respect to time is given as: [ λ is

Correct answer: C. λ e ^ - N = N _ 0 e ^ - N _ d = N _ 0 - N N _ d = N _ 0 (1 - e ^ - λ ) N _ d N _ 0 = f = 1 - e ^ - (df)/(dt) = λ e ^ -

JEE Main 202127 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Morning Shift
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2021 · 27 July · Morning Shift · Physics · Nuclear Physics

If f denotes the ratio of the number of nuclei decayed ( N _ d ) to...

If $f$ denotes the ratio of the number of nuclei decayed $\left({N}_{d}\right)$ to the number of nuclei at $t = 0, \left({N}_{0}\right) then for a collection of radioactive nuclei, the rate of change of$ f $with respect to time is given as:</p><p>$ [ \lambda $is the radioactive decay constant]</p>$

Options

A $- \lambda\left(1 - {e}^{- \lambdat}\right)$
B $\lambda\left(1 - {e}^{- \lambdat}\right)$
C $\lambda{e}^{- \lambdat}$
D $- \lambda{e}^{- \lambdat}$

Correct answer

C. $\lambda{e}^{- \lambdat}$

Step-by-step solution

N = {N}_{0}{e}^{- \lambdat}

{N}_{d} = {N}_{0} - N

{N}_{d} = {N}_{0}\left(1 - {e}^{- \lambda}\right)

\dfrac{{N}_{d}}{{N}_{0}} = f = 1 - {e}^{- \lambdat}

\dfrac{df}{dt} = \lambda{e}^{- \lambdat}

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