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

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12.Atoms

hard

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

$224$

$256$

$98$

$108$

Solution

(a) According to scattering formula

$N \propto \frac{1}{{{{\sin }^4}(\theta /2)}} \Rightarrow \frac{{{N_2}}}{{{N_1}}} = {\left[ {\frac{{\sin ({\theta _1}/2)}}{{\sin ({\theta _2}/2)}}} \right]^4}$

$ \Rightarrow \frac{{{N_2}}}{{{N_1}}} = {\left[ {\frac{{\sin \frac{{{{90}^o}}}{2}}}{{\sin \frac{{{{60}^o}}}{2}}}} \right]^4} = {\left[ {\frac{{\sin {{45}^o}}}{{\sin {{30}^o}}}} \right]^4}$

$ \Rightarrow {N_2} = {(\sqrt 2 )^4} \times {N_1} = 4 \times 56 = 224$

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