A fundamental particle of mass m and charge + | vedclass

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

Question

(A) At the distance of closest approach $(d)$,the entire initial kinetic energy of the particle is converted into electrostatic potential energy.Initi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) At the distance of closest approach $(d)$,the entire initial kinetic energy of the particle is converted into electrostatic potential energy.Initial kinetic energy $K = \frac{1}{2}mv^2$.Electrostatic potential energy $U = \frac{1}{4\pi \varepsilon_0} \cdot \frac{(e)(Ze)}{d}$.Equating $K = U$:$\frac{1}{2}mv^2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Ze^2}{d}$.Solving for $d$:$d = \frac{Ze^2}{2\pi \varepsilon_0 mv^2}$.

$A$ fundamental particle of mass $m$ and charge $+e$ is projected towards a much heavier charged particle of charge $+Ze$,where $Z > 0$. The distance of closest approach of the incident particle is ........

Similar Questions

Are atoms divisible or indivisible? Explain.

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:

An alpha particle is accelerated through a potential difference of $10^6 \ V$. Its kinetic energy will be $MeV$.

Find the number of atoms in $39.4 \ g$ of gold. The molar mass of gold is $197 \ g \ mol^{-1}$.

In the Rutherford experiment,$\alpha$-particles are scattered as shown in the figure. Which of the four paths is not possible?

Vedclass Test Series

Exam Paper Generator

Online Exam Module