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Find the decay constant (per minute) of a radioactive sample if at any instant its disintegration rate is 5000 disintegrations per minute and after 5 minutes, the rate is 1250 disintegrations per minute.

A

`0.4 ln 2`

B

`0.2 ln 2`

C

`0.1 ln 2`

D

`0.8 ln 2`

Text Solution

Verified by Experts

The correct Answer is:
A
According to question,
Rate of di sin tegration = Activity
`= (d N _(0))/( d t) =- N _(0) lamda = 5000......(1)`
Rate of di sin tegration = Activity
`= ( d N )/(dt) = - N lamda = 1250 .....(2)`
After time (t) = 5 min
`N = N _(0) e ^(- lamda t )`
equation `( 1) div (2)`
`(5000)/(1250) = (N _(0))/(N) = (N _(0))/( N _(0) e ^(- lamda t))`
` e ^( - lamda t ) = (1)/(4)`
`lamda t = 2 ln 2`
`lamda = ( 2 ln 2)/(t) = ( 2 ln 2)/(5) = 4 ln 2 min^(-1)`
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Knowledge Check

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

    A
    `0.4 in 2`
    B
    `0.2 in 2`
    C
    `0.1 in 2`
    D
    `0.8 in 2`

  • A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

    A
    `0.2 log_(e) 2`
    B
    `0.5 log_(e) 2`
    C
    `0.6 log_(e) 2`
    D
    `0.8 log_(e) 2`

  • Rate of disintegration per atom is called

    A
    decay constant
    B
    Mean life
    C
    Half life
    D
    none of these

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