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(A) According to Rutherford's scattering formula,the number of scattered particles $N$ at an angle $	heta$ is given by $N \propto \frac{1}{\sin^4(	h

Vedclass · Accepted Answer

$112/min, 12.5/min$

Vedclass · Answer

(A) According to Rutherford's scattering formula,the number of scattered particles $N$ at an angle $	heta$ is given by $N \propto \frac{1}{\sin^4(	heta/2)}$.Given $N_0 = 28$ per min at $	heta_0 = 90^o$.For $	heta_1 = 60^o$,$	heta_1/2 = 30^o$. Thus,$N_1 = N_0 	imes \frac{\sin^4(45^o)}{\sin^4(30^o)} = 28 	imes \frac{(1/\sqrt{2})^4}{(1/2)^4} = 28 	imes \frac{1/4}{1/16} = 28 	imes 4 = 112$ per min.For $	heta_2 = 120^o$,$	heta_2/2 = 60^o$. Thus,$N_2 = N_0 	imes \frac{\sin^4(45^o)}{\sin^4(60^o)} = 28 	imes \frac{(1/\sqrt{2})^4}{(\sqrt{3}/2)^4} = 28 	imes \frac{1/4}{9/16} = 28 	imes \frac{4}{9} \approx 12.44 \approx 12.5$ per min.Therefore,the values are $112/min$ and $12.5/min$.

If in Rutherford's experiment,the number of particles scattered at $90^o$ angle are $28$ per min,then the number of scattered particles at an angle $60^o$ and $120^o$ will be:

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