The ratio of momenta of an electron and an al | vedclass

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

Question

(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by the relation $p = \sqrt{2mqV}$,where $

Vedclass · Accepted Answer

$\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

Vedclass · Answer

(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by the relation $p = \sqrt{2mqV}$,where $m$ is the mass and $q$ is the charge of the particle.For an electron,$m = m_{e}$ and $q = e$.For an $\alpha$-particle,$m = m_{\alpha}$ and $q = 2e$.Taking the ratio of the momenta:$\frac{p_{e}}{p_{\alpha}} = \frac{\sqrt{2 m_{e} e V}}{\sqrt{2 m_{\alpha} (2e) V}}$$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{2 m_{e} e V}{4 m_{\alpha} e V}}$$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

Similar Questions

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

An $\alpha$-particle and a proton are accelerated from rest by the same potential,then the ratio of their de-Broglie wavelength is

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.

The wavelength of a very fast-moving electron $(v \approx c)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module