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(D) According to Rutherford's $\alpha$-particle scattering formula,the number of particles $N(	heta)$ scattered at an angle $	heta$ is given by $N(\

Vedclass · Accepted Answer

$28$

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(D) According to Rutherford's $\alpha$-particle scattering formula,the number of particles $N(	heta)$ scattered at an angle $	heta$ is given by $N(	heta) \propto \frac{1}{\sin^4(	heta/2)}$.Given $N(90^o) = 63$ and $	heta_1 = 90^o$,$	heta_2 = 120^o$.We have $\frac{N(120^o)}{N(90^o)} = \frac{\sin^4(90^o/2)}{\sin^4(120^o/2)} = \frac{\sin^4(45^o)}{\sin^4(60^o)}$.Substituting the values: $\sin(45^o) = \frac{1}{\sqrt{2}}$ and $\sin(60^o) = \frac{\sqrt{3}}{2}$.$\frac{N(120^o)}{63} = \frac{(1/\sqrt{2})^4}{(\sqrt{3}/2)^4} = \frac{1/4}{9/16} = \frac{1}{4} 	imes \frac{16}{9} = \frac{4}{9}$.$N(120^o) = 63 	imes \frac{4}{9} = 7 	imes 4 = 28$.

In the gold foil experiment,the number of deflected $\alpha$-particles at an angle of $90^o$ is $63$. What is the number of $\alpha$-particles deflected at $120^o$?

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