Question

At a certain time a radioactive sample contains 2 *(10)20 atoms and its disintegration rate is3*(10)10 atoms /second when 2*(10)15 atoms are left to decay its disintegration rate will be

Answer 1

Answer:

The disintegration rate is $3\times 10^{5}$ atoms/sec

Explanation:

If N are the number of atoms, disintegration rate R is given by

$\boxed{R=-\frac{dN}{dt} =-\lambda N}$

Thus,

$R_1=-\lambda N_1$

And

$R_2=-\lambda N_2$

∴ $\frac{R_2}{R_1}=\frac{N_2}{N_1}$

$\implies R_2=\frac{N_2}{N_1}\times R_1$

$\implies R_2=\frac{2\times 10^{15}}{2\times 10^{20}}\times 3\times 10^{10}$

$\implies R_2=3\times 10^{5}$ atoms/sec

Hope this helps.