Question

In Rutherford scattering experiment, the number of a-particles scattered at 60ŗ is 5 × 10⁶. The number of a-particles scattered at 120ŗ will be-

• A. 15 × 10⁶
• B. $\frac{3}{5}$× 10⁶
• C. $\frac{5}{9}$× 10⁶
• D. None

Answer 1

Answer: (C)

$\frac{5}{9}$× 10⁶

Answer 2

N µ $\frac{1}{\sin^{4}\frac{\theta}{2}}$; $\frac{N_{2}}{N_{1}}$ = $\frac{\sin^{4}\frac{\theta_{1}}{2}}{\sin^{4}\frac{\theta_{2}}{2}}$

or $\frac{N_{2}}{5 \times 10^{6}}$ = $\frac{\sin^{4}\frac{60º}{2}}{\sin^{4}\frac{120º}{2}}$ or $\frac{N_{2}}{5 \times 10^{6}}$=$\frac{\sin^{4}30º}{\sin^{4}60º}$

or N₂ = 5 × 10⁶ × $\left( \frac { 1 } { 2 } \right) ^ { 4 }$ $\left( \frac{2}{\sqrt{3}} \right)^{4}$ = $\frac{5}{9}$× 10⁶