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The weight of 1 curie .(82)Pb^(214) (t(1...

The weight of `1` curie `._(82)Pb^(214) (t_(1//2)=26.8 min)` in grams is

A

`3.1xx10^(-8) g`

B

`1.55xx10^(-8) g`

C

`6.2xx10^(-8)g`

D

`3.1xx10^(-10)g`

Text Solution

Verified by Experts

The correct Answer is:
a
We know that, `(-dN)/(dt) = lambda(6.023xx10^(23)xxw)/(M)`
Here, `(-dN)/(dt)` = 1 curie `= 3.7xx10^(10)dps`
`lambda = (0.693)/(t_(1//2)) = (0.693)/(26.8xx60)s^(-1)`
`:. 3.7xx10^(10) = (0.693)/(26.8xx60)xx(6.023xx10^(23)xxw)/(214)`
or `w = 3.1xx10^(-8)g`
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