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Question

(A) The number of scattered particles $N$ is related to the scattering angle $	heta$ by the Rutherford scattering formula: $N \propto \frac{1}{\sin^4

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(A) The number of scattered particles $N$ is related to the scattering angle $	heta$ by the Rutherford scattering formula: $N \propto \frac{1}{\sin^4(	heta/2)}$.Given $N_1 = 56$ at $	heta_1 = 90^\circ$,we need to find $N_2$ at $	heta_2 = 60^\circ$.Using the ratio: $\frac{N_2}{N_1} = \left[ \frac{\sin(	heta_1/2)}{\sin(	heta_2/2)} ight]^4$.Substituting the values: $\frac{N_2}{56} = \left[ \frac{\sin(90^\circ/2)}{\sin(60^\circ/2)} ight]^4 = \left[ \frac{\sin(45^\circ)}{\sin(30^\circ)} ight]^4$.Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$,we have $\frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Therefore,$\frac{N_2}{56} = (\sqrt{2})^4 = 4$.$N_2 = 56 	imes 4 = 224$.

If the number of alpha particles scattered at an angle of $90^\circ$ is $56$,how many particles will be scattered at an angle of $60^\circ$?

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