If the number of scattered alpha particles is | vedclass

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

Question

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

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(A) According to the Rutherford alpha-particle scattering formula,the number of scattered particles $N$ at an angle $	heta$ is given by $N \propto \frac{1}{\sin^4(	heta/2)}$.Therefore,we can write the ratio as: $\frac{N_2}{N_1} = \left[ \frac{\sin(	heta_1/2)}{\sin(	heta_2/2)} ight]^4$.Given: $N_1 = 56$,$	heta_1 = 90^o$,and $	heta_2 = 60^o$.Substituting the values: $\frac{N_2}{56} = \left[ \frac{\sin(90^o/2)}{\sin(60^o/2)} ight]^4 = \left[ \frac{\sin(45^o)}{\sin(30^o)} ight]^4$.Since $\sin(45^o) = \frac{1}{\sqrt{2}}$ and $\sin(30^o) = \frac{1}{2}$,we have: $\frac{N_2}{56} = \left[ \frac{1/\sqrt{2}}{1/2} ight]^4 = (\sqrt{2})^4 = 4$.Thus,$N_2 = 4 	imes 56 = 224$.

If the number of scattered alpha particles is $56$ at a scattering angle of ${90^o}$,then the number of scattered particles at a ${60^o}$ angle will be:

Similar Questions

How is an atom represented? Why?

Which source is associated with a line emission spectrum?

Assertion $(A)$: The impact parameter for $\alpha$-particles scattered by $180^{\circ}$ is zero.
Reason $(R)$: Zero impact parameter means that the $\alpha$-particles tend to hit the centre of the nucleus.

$\alpha$-particles of energy $400 \, KeV$ are bombarded on a nucleus of $_{82}Pb$. In the scattering of $\alpha$-particles, what will be the minimum distance of approach from the nucleus?

In the alpha particle scattering experiment,if $v$ is the initial velocity of the particle,then the distance of closest approach is $d$. If the velocity is doubled,then the distance of closest approach changes to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module