An elementary particle of mass m and charge + | vedclass

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

Question

(A) Let the distance of closest approach be $r$. At this distance,the entire initial kinetic energy of the incident particle is converted into electro

Vedclass · Accepted Answer

$\frac{Ze^2}{2\pi \varepsilon_0 mv^2}$

Vedclass · Answer

(A) Let the distance of closest approach be $r$. At this distance,the entire initial kinetic energy of the incident particle is converted into electrostatic potential energy.According to the law of conservation of energy:Initial Kinetic Energy = Potential Energy at distance $r$$\frac{1}{2}mv^2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{(Ze)(e)}{r}$Rearranging the equation to solve for $r$:$r = \frac{2Ze^2}{4\pi \varepsilon_0 mv^2} = \frac{Ze^2}{2\pi \varepsilon_0 mv^2}$

An elementary particle of mass $m$ and charge $+e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$ What is the closest possible approach of the incident particle?

Similar Questions

An $\alpha$-particle of $5 \; MeV$ energy strikes a stationary uranium nucleus at a scattering angle of $180^o$. The distance of closest approach of the $\alpha$-particle to the nucleus will be of the order of:

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

The diagram shows the path of four $\alpha$-particles of the same energy being scattered by the nucleus of an atom. Which of these is/are not physically possible?

What is a head-on collision? For that,state the impact parameter.

Atoms of the same element with different mass numbers are known as $\qquad$ .

Vedclass Test Series

Exam Paper Generator

Online Exam Module