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In a radioactive material the activity a...

In a radioactive material the activity at time `t_(1)` is `R_(1)` and at a later time `t_(2)`, it is `R_(2)`. If the decay constant of the material is `lambda`, then

A

`R _(1) = R_(2)`

B

`R _(1) = R _(2) e ^(-lamda ( t _(1) - t _(2)))`

C

`R _(1) = R _(2) e ^(lamda( t _(1) - t _(2)))`

D

`R _(1) = R _(2) (t _(2) //t _(1))`

Text Solution

Verified by Experts

The correct Answer is:
B
According to activity law, `R = R (0) e ^(-lamdar)`
`R _(1) = R _(0) e ^(-lamda t_(1)) and R _(2) = R _(0) e ^(- lamda t _(2))`
`(R _(1))/(R _(2)) = (R _(0) e ^(- lamda t _(1)))/( R _(0) e ^(- lamda t _(2))) = e ^(- lamda ( t _(1) - t_(2)))`
`or R _(1) = R _(2) e ^(- lamda (t _(1) - t _(2))`
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Knowledge Check

  • In a radioactive material the activity at time t_1 is R_1 and at a later time t_2 , it is R_1 . If the decay constant of the material is lamda , then

    A
    `R_1 = R_2 e^(-lamda(t_1 - t_2))`
    B
    `R_1 = R_2 e^(lamda(t_1 - t_2))`
    C
    `R_1 = R_2(t_2//t_1)`
    D
    `R_1 = R_2`

  • Disintegration constant of a radioactive material is lambda :

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    Its half lif eequal to `(log_(e) 2)/(lambda)`
    B
    Its mean life equal to `1/lambda`
    C
    At time equal to mean life, 63% of the initial radioactive material is left undecayed
    D
    After 3-half lives, `1/3` rd of the initial radioactive material is left undecayed.

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    C
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    D
    `R_(2)=R_(1) e^(lambda(t_(2)-t_(1)))`

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