A radioactive material decays by simultaneous emission of two particles of half lives 1620 and 810 years. The time (in years) after which one-fourth of the material remains, is

Options

A. 1080
B. 2430
C. 3240
D. 4860

Answer

A

Explanation

`(-dN)/(dt) = lamda_(1) N + lamda_(2) N`
`rArr ln ((N)/(N_(0)))= - (lamda_(1) + lamda_(2))t`
`N_(0) rarr` Initial number of atoms
`lamda_(1)= (0.693)/(1620) and lamda_(2)= (0.693)/(810)`
`(N)/(N_(0))= (1)/(4) rArr ln ((1)/(4)) = -((0.693)/(1620) + (0.693)/(810))t`
t= 1080 years