Two radioactive materials A & B have decay constant `3lamda` and `2lamda` respectively. At `t=0` the numbers of nuclei of A and B are `4N_(0)` and `2N_(0)` respectively then,

Options

A. Their number of radioactive nuclei will be equal at `t=(ln2)/(lamda)`
B. Their decay rate will be equals at `t=(ln4)/(lamda)`
C. Their decay rate will be equal at `t=(ln3)/(lamda)`
D. At `t=(ln4)/(lamda)` the decay rate of A will be greater than that of B.

Answer

A, C

Explanation

When no. of nuclei or equal `4N_(0)e^(-3lamdat)=2N_(0)e^(2lamdat)`
`impliest=(ln2)/(lamda)` when decay rate equal
`3lamda4N_(0)e^(-3It)=2lamda2N_(0)e^(-2lamdat)`
`t=(ln3)/(lamda)`
at `t=(ln4)/(lamda)` decay rate of `A=(3lamda)(4N_(0))e^(-3lamda(ln4)/(lamda))`
at `t=(ln4)/(lamda)` decay rate of `B=2lamda4N_(2)e^(-2lamda(ln4)/(lamda)`