The half-life period of RaB`(._(82)Pb^(214))` is `26.8 min`. The mass of one curie of RaB is

Options

A. `3.71 xx 10^(10)g`
B. `3.71 xx 10^(-10)g`
C. `8.61 xx 10^(10)g`
D. `3.064 xx 10^(-8)g`

Answer

d

Explanation

Here, `T=26.8 min=26.8 xx60 s`
:. Decay constant,
`lambda =(0.693)/(T)=(.693)/(26.8xx60)=4.32 xx10^(-4) s^(-1)`
Now, `1` curie is equal to `3.71 xx10^(10)` disintegrations per second `=3.71 xx 10^(10)`
If N be the number of atoms in one curie, then
`-(dN)/(dt) = lambda N`
or `3.71 xx10^(10) =431 xx 10^(-4)N`
`:. N=(3.71 xx10^(10))/(4.31 xx10^(-4))=8.607 xx 10^(13)`
Further, atomic weight `RaB=214` Avogardo's number `=6.025 xx 10^(23)`
Mass of one atoms `=(214)/(6.025 xx 10^(23))`
Mass of `N` atoms` =((214)/(6.025 xx 10^(23))) xx(8.607 xx 10^(23))`
`=3.064 xx 10^(-8) g` .