Alpha-particle scattering experiment

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Question

If scattering particles are 56 for 90^o angle then find the number of particles at 60^o angle.

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Solution

According to scattering formula
$N \propto \frac{1}{\sin^{4} (θ / 2)} \Rightarrow \frac{N_{2}}{N_{1}} = {[\frac{\sin (θ_{1} / 2)}{\sin (θ_{2} / 2)}]}^{4}$
$\Rightarrow \frac{N_{2}}{N_{1}} = {[\frac{\sin \frac{90^{\circ}}{2}}{\sin \frac{60^{\circ}}{2}}]}^{4} = {[\frac{\sin 45^{\circ}}{\sin 30^{\circ}}]}^{4} \Rightarrow N_{2} = (\sqrt{2})^{4} \times N_{1} = 4 \times 56 = 224$

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Alpha-particle scattering experiment

Remember concepts with our Masterclasses.

If scattering particles are 56 for 90o angle then find the number of particles at 60o angle.

According to scattering formulaN∝1sin4⁡(θ/2)⇒N2N1=sin⁡θ1/2sin⁡θ2/24⇒N2N1=sin⁡90∘2sin⁡60∘24=sin⁡45∘sin⁡30∘4 ⇒N2=(2)4×N1=4×56=224

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If scattering particles are 56 for 90^o angle then find the number of particles at 60^o angle.