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

Options

A. 1500 yr
B. 300 yr
C. 449 yr
D. 810 yr

Answer

C

Explanation

`lamda_(alpha)=(1)/(1620)` per year and `lamda_(beta)=(1)/(405)` per year and it is given that the fraction of the remained activity `(R_(1))/(R_(2))=1/4`
Total decay constant,
`lamda=lamda_(alpha)+lamda_(beta)=(1)/(1620)+(1)/(405)=(1)/(324)` per year
We knew that, `R=R_(0)e^(-lamdat)impliest=(1)/(lamda)log"(R_(0))/(R)`
`impliest=(1)/(lamda)log""_(e)2=324xx2xx0.693=449yr`