A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is

  • Option 1)

    0.4 ln 2

  • Option 2)

    0.2 ln 2

  • Option 3)

    0.1 ln 2

  • Option 4)

    0.8 ln 2

 

Answers (1)

As we learnt in

Number of nuclei after disintegration -

N=N_{0}e^{-\lambda t} or A=A_{0}e^{-\lambda t}

- wherein

Number of nucleor activity at a time is exponentional function

 

 A=A_{0}e^{-\lambda t}     or      A=\frac{A_{0}}{2^{t/t_{1/2}}}

A = 1250

A0 = 5000

\Rightarrow\ \;1250=\frac{5000}{2^{t/t_{1/2}}}    or     2^{\frac{t}{t_{1/2}}}=4

t=2t_{\frac{1}{2}}

t = 5 minutes

\therefore\ \;t_{\frac{1}{2}}=2.5\ minutes

\lambda=\frac{l_{n}^{2}}{t_{\frac{1}{2}}}=0.4\ l_{n}^{2}

Correct option is 1.


Option 1)

0.4 ln 2

This is the correct option.

Option 2)

0.2 ln 2

This is an incorrect option.

Option 3)

0.1 ln 2

This is an incorrect option.

Option 4)

0.8 ln 2

This is an incorrect option.

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