A radioactive material has mean lives of 1620 yr and 520 yr for `alphaandbeta-"emission"`, respectively. The material decays by simultaneous `alphaandbeta-"emissions`. The time in which 1/4th of the material remains intact is

Options

A. 4.675 yr
B. 7.20 yr
C. 5.45 yr
D. 324 yr

Answer

C

Explanation

`lamda=(lamda_(1)+lamda_(2))=(1)/(1620)+(1)/(520)=2.54xx10^(-3)yr^(-1)`
`therefore" " t_(1//2)=(1n(2))/(lamda)=272.8 yr`
(1/4) th of the material remainsintact after 2 half-lives.