If the number of alpha particles scattered at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The number of scattered particles $N$ is related to the scattering angle $	heta$ by the Rutherford scattering formula: $N \propto \frac{1}{\sin^4

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(A) The number of scattered particles $N$ is related to the scattering angle $	heta$ by the Rutherford scattering formula: $N \propto \frac{1}{\sin^4(	heta/2)}$.Given $N_1 = 56$ at $	heta_1 = 90^\circ$,we need to find $N_2$ at $	heta_2 = 60^\circ$.Using the ratio: $\frac{N_2}{N_1} = \left[ \frac{\sin(	heta_1/2)}{\sin(	heta_2/2)} ight]^4$.Substituting the values: $\frac{N_2}{56} = \left[ \frac{\sin(90^\circ/2)}{\sin(60^\circ/2)} ight]^4 = \left[ \frac{\sin(45^\circ)}{\sin(30^\circ)} ight]^4$.Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$,we have $\frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Therefore,$\frac{N_2}{56} = (\sqrt{2})^4 = 4$.$N_2 = 56 	imes 4 = 224$.

If the number of alpha particles scattered at an angle of $90^\circ$ is $56$,how many particles will be scattered at an angle of $60^\circ$?

Similar Questions

Based on which experiment did the Rutherford nuclear model come from?

According to the Rutherford's atomic model,the electrons inside the atom are

According to Richard Feynman,matter is made of which constituent particles? Why is it important? Write the atomic hypothesis.

Who discovered the spin quantum number?

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module