In a radioactive material, a fraction of active material remaining after the time tis(9)/(16) . The fraction that was remaining after the time (t)/(2) is :

Correct answer: C. (3)/(4) N = N _ 0 e ^ - . . . (1) N ' = N _ 0 e ^ - ( )/(2) . (2) from (1) & (2) ( N ' N _ 0 ) = ( N N _ 0 )(1)/(2) = ((9)/(16))(1)/(2) = (3)/(4).

JEE Main 20203 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Morning Shift
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2020 · 3 September · Morning Shift · Physics · Nuclear Physics

In a radioactive material, a fraction of active material remaining...

In a radioactive material, a fraction of active material remaining after the time

tis\dfrac{9}{16}

. The fraction that was remaining after the time

\dfrac{t}{2}

is :

Options

A $\dfrac{4}{5}$
B $\dfrac{3}{5}$
C $\dfrac{3}{4}$
D $\dfrac{7}{8}$

Correct answer

C. $\dfrac{3}{4}$

Step-by-step solution

$N = {N}_{0}{e}^{- \lambdat} . . .$ \left(1\right)

$N ' = {N}_{0}{e}^{- \dfrac{\lambdat}{2}} \ldots . \left(2\right)$

from (1) & (2)

\left(\dfrac{N '}{{N}_{0}}\right) = \left(\dfrac{N}{{N}_{0}}\right)\dfrac{1}{2} = \left(\dfrac{9}{16}\right)\dfrac{1}{2} = \dfrac{3}{4}.

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