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Q.

In Rutherford experiments on $α$ -ray scattering, the number of particles scattered at 90^o be 28 per minute. Then the number of particles scattered per minute by the same foil, but at 60^o is:

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a

56

b

112

c

60

d

120

answer is B.

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Detailed Solution

The number of α –particles scattered at an angle θ is given by:
$\begin{array}{l} N = \frac{{Qntz}^{2} e^{4}}{{(8 {πε}_{0})}^{2} r^{2} E^{2} (\sin θ / 2)^{4}}; N \propto {(\frac{1}{\sin θ / 2})}^{4} \\ \Rightarrow \frac{N_{1}}{N_{2}} = {(\frac{Sin (\frac{θ_{2}}{2})}{\sin (\frac{θ_{1}}{2})})}^{4}; \frac{28}{N_{2}} = \frac{\sin^{4} (\frac{60}{2})}{\sin^{4} (\frac{90}{2})} = \frac{\sin^{4} 30}{\sin^{4} 45} \\ ∴ N_{2} = 112 \end{array}$

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