The half life of a radioactive substance is 20 minutes. The approximate time interval (t_2-t_1) between the time t_2 when (2)/(3) of it has decayed and time t_1 and (1)/(3) of it had decayed is :

Correct answer: B. 20 ~min t_ (1)/(2) =20 minutes N=N_0 e^ -λ t_2 λ t_1= 3 (2)/(3) ~N_0=N_0 e^ -λ t_2 t_1=(1)/(λ) 3 (2)/(3) ~N_0=N_0 e^ -λ t_2 t_2=(1)/(λ) (3)/(2) t_2-t_1=(1)/(λ) [ (3)/(2)- 3] =(1)/(λ) [(1)/(2)]=(0.693)/(λ) =20

JEE Main 2011

Exam: JEE Main
Year: 2011
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2011 · Physics · Nuclear Physics

The half life of a radioactive substance is 20 minutes. The...

The half life of a radioactive substance is 20 minutes. The approximate time interval

\left(t_2-t_1\right)

between the time

t_2

when

\frac{2}{3}

of it has decayed and time

t_1 and \frac{1}{3}

of it had decayed is :

Options

A $14 \mathrm{~min}$
B $20 \mathrm{~min}$
C $28 \mathrm{~min}$
D $7 \mathrm{~min}$

Correct answer

B. $20 \mathrm{~min}$

Step-by-step solution

\mathrm{t}_{\frac{1}{2}}=20

minutes

\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_2} \quad \lambda \mathrm{t}_1=\ln 3

\frac{2}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_2} \mathrm{t}_1=\frac{1}{\lambda} \ln 3

\frac{2}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda t_2}

\mathrm{t}_2=\frac{1}{\lambda} \ln \frac{3}{2}

\mathrm{t}_2-\mathrm{t}_1=\frac{1}{\lambda}

\left[\ln \frac{3}{2}-\ln 3\right] =\frac{1}{\lambda} \ln \left[\frac{1}{2}\right]=\frac{0.693}{\lambda} =20 \min

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