Question

If 1 g of ${}^{226}Ra$ emits $11.6\times {10}^{17}\alpha$-particles in a years. If $\lambda =4.358\times {10}^{-4}y{r}^{-1}$, calculate value of Avogadro's number?

• A. $N_0=6.02\times {10}^{23}mo{l}^{-1}$
• B. $N_0=2.32\times {10}^{23}mo{l}^{-1}$
• C. $N_0=6.22\times {10}^{23}mo{l}^{-1}$
• D. None of these

Answer 1

The correct option is A $N_0=6.02\times {10}^{23}mo{l}^{-1}$

1 g radium $=\frac{N_O}{226}atoms$
$-\frac{dN}{dt}=\frac{\lambda N_O}{226}$
$11.6\times {10}^{17}=\frac{4.353\times {10}^{-4}{N}_0}{226}$
$N_0=6.02\times {10}^{23}mo{l}^{-1}$