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Question

In Rutherford experiments of $$\alpha  - ray$$ scattering the number of particles scattered at $${90^0}$$ be $$28$$ per minute$$.$$ Then the number of particles scattered per minute by the same foil but at $${60^0}$$ are 


A
56
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B
112
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C
60
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D
120
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Solution

The correct option is D $$112$$
No. of particles scattered at angle is 
$$N \propto \dfrac{1}{sin^4 \left( \dfrac{\theta}{2} \right)} $$
Given:- the number of particles scattered at $$90^o$$ be $$28$$ per minute
$$\Rightarrow $$ $$\dfrac{N_1}{N_2} = \dfrac{sin^4 \left( \dfrac{\theta_2}{2} \right) }{sin^4 \left( \dfrac{\theta_1}{2} \right)}$$

$$\Rightarrow $$ $$\dfrac{28}{N_2} = \dfrac{sin^4 \left( \dfrac{60^0}{2} \right) }{sin^4 \left( \dfrac{90^o}{2} \right)}$$

$$\Rightarrow $$ $$N_2 = 112$$

Physics

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