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12.Atoms
medium
In Rutherford's experiment, number of particles scattered at $90^{\circ}$ angle are $x$ per second. Number particles scattered per second at angle $60^{\circ}$ is
A
$x$
B
$4 x$
C
$8 x$
D
$16 x$
Solution
(b)
$N(\theta) \propto \frac{Z^2}{\sin ^4\left(\frac{\theta}{2}\right) E^2}$
$\text { or } N (\theta) \propto \frac{1}{\sin ^4\left(\frac{\theta}{2}\right)}$
$\theta=60^{\circ}$ and $90^{\circ}$
$\therefore \frac{N(90)}{N(60)} \propto \frac{\sin ^4 30}{\sin ^4 45}$
$N\left(90^{\circ}\right)=x$
$N\left(60^{\circ}\right)=4 x$
Standard 12
Physics
