In Rutherford’s $α$ -particle scattering experiment with their gold foil 8100 scintillations per minute are observed at an angle of 60^o. The number of scintillations per minute at 120^o will be:

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2025

100

32400

900

answer is D.

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Detailed Solution

The number of α –particles scattered at an angle θ is given by
$\begin{array}{l} N = \frac{{Qntz}^{2} e^{4}}{{(8 {πε}_{0})}^{2} r^{2} E^{2} (\sin θ / 2)^{4}}; N \propto \frac{1}{{Sin}^{4} (\frac{θ}{2})} \\ \Rightarrow \frac{N_{1}}{N_{2}} = \frac{{Sin}^{4} (\frac{θ_{2}}{2})}{{Sin}^{4} (\frac{θ_{1}}{2})} \end{array}$
$\begin{array}{l} \Rightarrow \frac{8100}{N_{2}} = \frac{{Sin}^{4} (\frac{120}{2})}{{Sin}^{4} (\frac{60}{2})} = \frac{{Sin}^{4} (60 °)}{{Sin}^{4} (30 °)} \\ ∴ N_{2} = \frac{8100}{9} = 900 \end{array}$