In the alpha particle scattering experiment,i | vedclass

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

Question

(D) The distance of closest approach $r_0$ is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential e

Vedclass · Accepted Answer

$\frac{d}{4}$

Vedclass · Answer

(D) The distance of closest approach $r_0$ is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy at the point of closest approach:$\frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$From this equation,we can see that $v^2 \propto \frac{1}{r_0}$,which implies $r_0 \propto \frac{1}{v^2}$.Given that the initial distance of closest approach is $d$ for velocity $v$,let the new distance be $d'$ for velocity $2v$.$\frac{d'}{d} = \frac{v^2}{(2v)^2} = \frac{v^2}{4v^2} = \frac{1}{4}$Therefore,$d' = \frac{d}{4}$.

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:

Similar Questions

We consider the Thomson model of the hydrogen atom in which the proton charge is distributed uniformly over a spherical volume of radius $R = 0.25 \,\mathring A$. Applying the Bohr quantization condition in this model,the ground state energy (in $eV$) of the electron will be close to:

What is shown by Thomson's experiments of electric discharge through gases? And explain the plum pudding model.

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

Which source is associated with a line emission spectrum?

How is an atom represented? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module