Question

In a Rutherford experiment, the number of particles per scattered at $ 90^\circ $ angle are 28 per minute, then the number of scattered particles at an angle $ 60^\circ $ and $ 120^\circ $ are
(A) 117 per minute, 25 per minute
(B) 50 per minute, 12.5 per minute
(C) 100 per minute, 200 per minute
(D) 112 per minute, 12.4 per minute

Answer 1

Hint : The number of particles scattered per unit time is inversely proportional to the fourth power of the sine of half of the scattering angle. We need to find the proportionality constant using the given values.

Formula used: In this solution we will be using the following formula;
 $ N(\theta ) = \dfrac{k}{{{{\sin }^4}\dfrac{\theta }{2}}} $ where $ N(\theta ) $ is the number of particle scattered per unit time, $ k $ is a constant, and $ \theta $ is the scattering angle.

Complete step by step answer:
We are given the number of particles scattered at a particular angle, and we are told to find the number of particles scattered at other angles.
To solve this, we must note that the number of particles scattered at a particular angle is inversely proportional to the fourth power of the sine of the angle. In mathematical terms, it is written as
 $ N(\theta ) \propto \dfrac{1}{{{{\sin }^4}\dfrac{\theta }{2}}} $
 $ \Rightarrow N(\theta ) = \dfrac{k}{{{{\sin }^4}\dfrac{\theta }{2}}} $ where $ N(\theta ) $ is the number of particle scattered, $ k $ is a proportionality constant, and $ \theta $ is the scattering angle.
Then, from given values in the question, we have that
 $ 28 = \dfrac{k}{{{{\sin }^4}\dfrac{{90^\circ }}{2}}} = \dfrac{k}{{0.25}} = 4k $ since $ \sin 45^\circ = \dfrac{{\sqrt 2 }}{2} $
Hence,
 $ k = 7 $
Now, using the known constant, we can find the number of particles scattered per unit time for angle 60 and 120 degrees.
 $ N(60) = \dfrac{7}{{{{\sin }^4}\dfrac{{60}}{2}^\circ }} = \dfrac{7}{{{{\sin }^4}30^\circ }} $
By computing we have
 $ N(60) = 112{\min ^{ - 1}} $
Similarly, for angle 120 degrees we have
 $ N(120) = \dfrac{7}{{{{\sin }^4}\dfrac{{120}}{2}^\circ }} = \dfrac{7}{{{{\sin }^4}60^\circ }} $
By computation, we have that
 $ N(60) = 12.4{\min ^{ - 1}} $
Hence, the correct option is D .

Note:
Alternatively, for exam purposes, after calculating for number of particles per minute at 60 degrees to be 112 per minute, we could simply opt to choose D as the answer since no other options have the 112 per minute as the first answer.