Test Index

Atoms and Nuclei

Section: Physics

Question : 9 of 33

Marks: +1, -0

8\%

77\%

Solution:

When

8\%

of nuclei decays, then remaining nuclei are

92\%

from

As,

N=N_0 e^{-\lambda t}

Here,

N=0.92 N_0

So,

0.92 N_0=N_0 e^{-\lambda t_1}

When,

77\%

nuclei decay remaining number of nuclei are

23\%

So,

\;\; N=0.23 N_0

0.23 N_0=N_0 e^{-\lambda t_2}

By dividing in equation (i) by equation (ii)

\; \frac{0.92}{0.23} = \; \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{\lambda (t_2-t_1)}

So,

e^{\lambda (t_2-t_1)} = 4

By taking natural log of both sides

\lambda (t_2-t_1) \; = \ln 4

\; \text{ and } \; t_2-t_1 \; =12 \; \text{ minutes } \;

12 \lambda \; = 2 \ln 2

\lambda \; = \; \frac{\ln 2}{6}

and relation between half life and decay constant is

\lambda = \; \frac{\ln 2}{t_{1/2}}

So,

\; \frac{\ln 2}{6} = \; \frac{\ln 2}{t_{1/2}}

and

t_{1/2}=6

minutes

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