Question

the rate of electron emission from 4mg of 210Pb80 with half life 5 days is

answer is 1.84×10^13

Answer 1

The rate of electron emission $\dfrac{dN}{dt}$ is proportional to the number of nucleus $N$ in sample, accoring to the decay law (see https://en.wikipedia.org/wiki/Radioactive_decay#Universal_law):

$$\left| \dfrac{dN}{dt} \right| = \lambda N$$

where $\lambda = \dfrac{\ln2}{T}$, where $T = 5days = 432000s$ is the half life of 210Pb80.

The number of nuclei in sample with mass $m = 4mg = 0.004g$ is given as follows:

$$N = \dfrac{m}{M}\cdot N_A$$

where $M = 210\dfrac{g}{mol}$ is the molar mass of Pb, and $N_A \approx 6.022\times 10^{23}mol^{-1}$ is the Avogadro number. Putting it all together, obtain:

$$\left| \dfrac{dN}{dt} \right| = \dfrac{mN_A\cdot \ln2}{TM}\\ \left| \dfrac{dN}{dt} \right| = \dfrac{0.004g\cdot 6.022\times 10^{23}mol^{-1}\cdot \ln2}{432000s\cdot 210g/mol} \approx 1.84\times 10^{13}s^{-1}$$

Answer. $1.84\times 10^{13}s^{-1}$.