Question

1 g of ${}^{226}Ra$ emits $11.6\times {10}^{17}\alpha$-particles in a year. If $\lambda =4.353\times {10}^{-4}{years}^{-1}$, calculate the value of Avogadro's number.

• A. $N_0=6.02\times {10}^{23}mo{l}^{-1}$
• B. $N_0=6.66\times {10}^{23}mo{l}^{-1}$
• C. $N_0=2.02\times {10}^{23}mo{l}^{-1}$
• D. $N_0=4.32\times {10}^{23}mo{l}^{-1}$

Answer 1

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

1 g radium = $\frac{N_0}{226}$ atoms

$-\frac{dN}{dt}=\frac{\lambda N_o}{226}$

$11.6\times {10}^{17}=\frac{4.353\times {10}^{-4}{N}_o}{226}$

$\therefore N_0=6.02\times {10}^{23}mo{l}^{-1}$

Hence, option A is correct.