If scattering particles are 56 for {90^o} ang | vedclass

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Question

(a) According to scattering formula

$N \propto \frac{1}{{{{\sin }^4}(	heta /2)}} \Rightarrow \frac{{{N_2}}}{{{N_1}}} = {\left[ {\frac{{\sin ({	he

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(a) According to scattering formula

$N \propto \frac{1}{{{{\sin }^4}(	heta /2)}} \Rightarrow \frac{{{N_2}}}{{{N_1}}} = {\left[ {\frac{{\sin ({	heta _1}/2)}}{{\sin ({	heta _2}/2)}}} ight]^4}$

$ \Rightarrow \frac{{{N_2}}}{{{N_1}}} = {\left[ {\frac{{\sin \frac{{{{90}^o}}}{2}}}{{\sin \frac{{{{60}^o}}}{2}}}} ight]^4} = {\left[ {\frac{{\sin {{45}^o}}}{{\sin {{30}^o}}}} ight]^4}$

$ \Rightarrow {N_2} = {(\sqrt 2 )^4} 	imes {N_1} = 4 	imes 56 = 224$

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

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