Question

When a silver foil (Z=47) was used in an $\alpha$-ray scattering experiment, the number of $\alpha$ partied scattered at ${30}^{\circ }$ aws 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. $26$

Answer 1

Answer: (D)

$26$

No of alpha particles scattered by silver foil

$N_s\times \frac{{\left(Z_{s}e\right)}^2}{{\left(\sin \frac{\theta }{2}\right)}^4}...\left(1\right)$

$N_{al}\times \frac{{\left(Z_{al}e\right)}^2}{{\left(\sin \frac{\theta }{2}\right)}^4}....\left(2\right)$

$\frac{2}{1}=\frac{N_{al}}{N_s}=\frac{{\left(Z_{al}\right)}^2}{{\left(Z_s\right)}^2}or{N}_{al}=N_s\times \frac{{\left(Z_{al}\right)}^2}{{\left(Z_s\right)}^2}$

$N_{al}{=N}_s\times {\left(\frac{47}{13}\right)}^2[\because N_s=200]$

$\therefore N_{al}=200{\left(3.615\right)}^2=2620.8$