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

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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 ....... .

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