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(B) The number of alpha particles $N(	heta)$ scattered at an angle $	heta$ is given by the relation:$N(	heta) \propto \frac{1}{\sin^4(	heta/2)}$Gi

Vedclass · Accepted Answer

$4 x$

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(B) The number of alpha particles $N(	heta)$ scattered at an angle $	heta$ is given by the relation:$N(	heta) \propto \frac{1}{\sin^4(	heta/2)}$Given:$N(90^{\circ}) = x$We need to find $N(60^{\circ})$:$\frac{N(60^{\circ})}{N(90^{\circ})} = \frac{\sin^4(90^{\circ}/2)}{\sin^4(60^{\circ}/2)}$$\frac{N(60^{\circ})}{x} = \frac{\sin^4(45^{\circ})}{\sin^4(30^{\circ})}$Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$ and $\sin(30^{\circ}) = \frac{1}{2}$:$\frac{N(60^{\circ})}{x} = \frac{(1/\sqrt{2})^4}{(1/2)^4} = \frac{1/4}{1/16} = \frac{16}{4} = 4$Therefore,$N(60^{\circ}) = 4x$.

In Rutherford's experiment,the number of particles scattered at a $90^{\circ}$ angle is $x$ per second. The number of particles scattered per second at an angle of $60^{\circ}$ is:

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