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Question: If scattering particles are 56 for \(90^{o}\) angle then this will be at \(60^{o}\) angle...

If scattering particles are 56 for 90o90^{o} angle then this will be at 60o60^{o} angle

A

224

B

256

C

98

D

108

Answer

224

Explanation

Solution

Using Scattering formula

N∝1sin⁡4(θ/2)N \propto \frac{1}{\sin^{4}(\theta/2)}⇒ N2N1=[sin⁡(θ12)sin⁡(θ22)]4\frac{N_{2}}{N_{1}} = \left\lbrack \frac{\sin\left( \frac{\theta_{1}}{2} \right)}{\sin\left( \frac{\theta_{2}}{2} \right)} \right\rbrack^{4}

⇒N2N1=[sin⁡(90∘2)sin⁡(60∘2)]4=[sin⁡45∘sin⁡30∘]4=4\frac{N_{2}}{N_{1}} = \left\lbrack \frac{\sin\left( \frac{90{^\circ}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2} \right)} \right\rbrack^{4} = \left\lbrack \frac{\sin 45{^\circ}}{\sin 30{^\circ}} \right\rbrack^{4} = 4

⇒ N2=4N1=4×56=224N_{2} = 4N_{1} = 4 \times 56 = 224