A radioacive element decays by parallel path as given below:
`A overset(lambda_(1))(to) lambda_(1) = 1.8 xx 10^(-2) sec^(-1)`
`2A overset(lambda_(2))(to) B lambda_(2) = 10^(-3) sec^(-1)`
Average life of radio nuclide A will be:

Options

A. 52.63 sec
B. 500 sec
C. 50 sec
D. 120 sec

Answer

c

Explanation

`lambda = lambda_(1) + 2 lambda_(2)`
`= 1.80 xx 10^(-2) + 2 xx 10^(-3)`
`= 2 xx 10^(-2) sec^(-1)`
`tau = (1)/(lambda) = (1)/(2 xx 10^(-2)) = 50` sec