Question

$(-\frac{dN}{dt}=\lambda N)$ represents the rate of decay in disintegration of atoms per unit time, where $N$ is the number of atoms disintegrating at that time and $\lambda$ is the disintegration constant. Calculate the number of curies when number of atoms equal to one Avogadro's number disintegrates in the case when $\lambda =0.6146\times {10}^{-10}{s}^{-1}$

Answer 1

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

ATQ, when $N=6.02\times {10}^{23}atoms\&\lambda =0.6146\times {10}^{-10}{s}^{-1}\to -\frac{dN}{dt}=6.02\times {10}^{23}\times 0.6146\times {10}^{-10}=3.69\times {10}^{13}atoms/s\simeq 3.7\times {10}^{13}atoms/s$

Now, we know,

$1curie=3.7\times {10}^{10}atom/s\to 1atom/s=\frac{1}{3.7\times {10}^{10}}curieor3.7\times {10}^{13}atoms/s=d\frac{3.7\times {10}^{13}}{3.7\times {10}^{10}}curie=1000curie\therefore -\frac{dN}{dt}=1000curie$