Atomic StructureEasy
Question
In the Rutherford scattering experiment, the number of alpha particles scattered at an angle θ = 60o is 36 per minute. The number of alpha particles per minute scattered at angles θ = 90o is (Assume all other conditions to be identical)
Options
A.144
B.9
C.36
D.16
Solution
$N_{\alpha} \propto \frac{1}{\sin^{4}\left( \frac{\theta}{2} \right)}$
$\therefore\frac{N_{1}}{N_{2}} = \left\lbrack \frac{\sin\left( \frac{90}{2} \right)}{\sin\left( \frac{60}{2} \right)} \right\rbrack^{4} \Rightarrow \frac{36}{N_{2}} = \frac{4}{1} \Rightarrow N_{2} = 9$
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