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

(d) Here `T = 26.8` minutes
=`26.8 xx 60 sec`
`:.` Decay constant,
`lamda = (0.693)/(T) = (0.693)/(26.8 xx 60)`
=`4.32 xx 10^4 sec^-1`
Now `1` courie is equal to `3.71 xx 10^10` disintegrations per second
`:. lamda = 3.71 xx 10^10`
If `N` be the number of atoms in one curie, then
`- (dN)/(dt) = lamda N`
or `3.71 xx 10^10 = 431 xx 10^-4`
`:. N = (3.71 xx 10^10)/(4.31 xx 10^-4) = 8.607 xx 10^13`
Further, atomic weight of `RaB = 214` and Avogadro's number `= 6.025 xx 10^23`
Mass of one atom `=(214)/(6.025 xx 1023)`
Mass of `N` atoms
=`((214)/(6.025 xx 10^23)) xx (8.607 xx 10^13)`
=`3.064 xx 10^-8 g`.