A radioactive material decays by simulataneous emission of two particle from the with respective half - lives `1620` and `810` year . The time , in year , after which one - fourth of the material remains is

Options

A. 4860 yr
B. 3240 yr
C. 2340 yr
D. 1080 yr

Answer

C

Explanation

Let the initial number of atoms at time, `t=0` be `N_(0)`. Let N be the number of atoms at any instant t.
Mean life `tau=(1)/(lamda)`, where `lamda` is disintegration constant.
Given, `t=tau`
According to radioactive disintegration law, `N=N_(0)e^(-lamdat)`
or `N=N_(0)e^(-lamda xx(1)/(lamda))=(N_(0))/(e) or (N_(0))/(N)=e`