The number of `alpha`-particles emitted per second by 1g fo `.^(226)Ra` is `.7 xx 10^(10)`. The decay constant is:

Options

A. `1.39 xx 10^(-11) sec^(-1)`
B. `13.9 xx 10^(-11) sec^(-1)`
C. `139 xx 10^(-10) sec^(-1)`
D. `13.9 xx 10^(-10) sec^(-1)`

Answer

a

Explanation

`("No. of atoms disintegration per second")/("Total number of atoms present")`
or `(3.7 xx 10^(10))/((6.02 xx 10^(23))/(226)) = (226 xx 3.7 xx 10^(10))/(6.02 xx 10^(23)) = lambda`