An electron of mass m_{e} and a proton of mas | vedclass

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

Question

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,

Vedclass · Accepted Answer

$1836$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the speed.Given that both particles move with the same speed $v$,the ratio of their wavelengths is:$\frac{\lambda_{e}}{\lambda_{p}} = \frac{\frac{h}{m_{e}v}}{\frac{h}{m_{p}v}} = \frac{m_{p}}{m_{e}}$.Substituting the given relation $m_{p} = 1836 m_{e}$:$\frac{\lambda_{e}}{\lambda_{p}} = \frac{1836 m_{e}}{m_{e}} = 1836$.

An electron of mass $m_{e}$ and a proton of mass $m_{p} = 1836 m_{e}$ are moving with the same speed. The ratio of their de Broglie wavelength $\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}}$ will be ....... .

Similar Questions

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

What is the de Broglie wavelength (in $\mathring{A}$) of an $\alpha$-particle accelerated through a potential difference of $V$ volts?

Find the ratio of the de Broglie wavelength of an oxygen molecule at $21^oC$ to that of a nitrogen molecule at $63^oC$.

If the kinetic energy of a free electron doubles,its de-Broglie wavelength $\lambda$ changes by a factor of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module