An electron and a proton are accelerated thro | vedclass

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

Question

(B) Both the electron and the proton are accelerated through the same potential difference $V$,so they acquire the same kinetic energy $K = eV$.The mo

Vedclass · Accepted Answer

$\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$

Vedclass · Answer

(B) Both the electron and the proton are accelerated through the same potential difference $V$,so they acquire the same kinetic energy $K = eV$.The momentum $P$ of a particle is related to its kinetic energy $K$ by the formula $P = \sqrt{2mK}$.For the electron,the momentum is $P_{e} = \sqrt{2m_{e}K}$.For the proton,the momentum is $P_{p} = \sqrt{2m_{p}K}$.The de-Broglie wavelength is given by $\lambda = \frac{h}{P}$.Therefore,the ratio of the wavelengths is $\frac{\lambda_{p}}{\lambda_{e}} = \frac{h/P_{p}}{h/P_{e}} = \frac{P_{e}}{P_{p}}$.Substituting the expressions for momentum: $\frac{\lambda_{p}}{\lambda_{e}} = \frac{\sqrt{2m_{e}K}}{\sqrt{2m_{p}K}} = \sqrt{\frac{m_{e}}{m_{p}}} = \left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$.

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

Similar Questions

An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .

The kinetic energy of an electron gets tripled,then the de-Broglie wavelength associated with it changes by a factor of:

The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

Vedclass Test Series

Exam Paper Generator

Online Exam Module