Question

When a silver foil (Z = 47) was used in a $\alpha$-ray scattering experiment, the number of $\alpha$ particles scattered at ${30}^{\circ }$ was found to be 200 per minute. If the silver foil is replaced by aluminium (z = 13) foil of same thickness, the number of $\alpha$ -particles scattered per minute at ${30}^{\circ }$ is nearly equal to

• A. 15
• B. 30
• C. 10
• D. 20

Answer 1

The correct option is A 15

$N\propto Z^2$
$\frac{N_2}{N_1}={\left[\frac{z_2}{z_1}\right]}^2={\left[\frac{13}{47}\right]}^2$
$\frac{N_2}{200}=\frac{13\times 13}{47\times 47}$
$N_2=\frac{13\times 13\times 200}{47\times 47}=15$