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The half-life of 4.0mg beta- emitter of ...

The half-life of `4.0mg beta-` emitter of `.^(210)X` is `5` day and the average energy of emitted `beta-` particle is `0.34MeV`. At what rate in watts does the sample emits energy?

A

`2.0`

B

`0.1`

C

`1.5`

D

`1.0`

Text Solution

Verified by Experts

The correct Answer is:
`(d)`
Power = Energy of `beta` in `J xx "No of" beta-` particles emitted/sec.
`= E xx (-(dN)/(dt)) = E xx lambda xx N`
`= 0.34xx10^(6)xx1.6xx10^(-19) xx (0.693)/(5xx24xx60xx60)`
`xx (4xx10^(-3)xx6.023xx10^(23))/(210) = 1J//sec = 1` watt
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