A photon and an electron are given the same e | vedclass

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(A) For a photon,the energy $E$ is given by $E = \frac{hc}{\lambda_{Ph}}$,so $\lambda_{Ph} = \frac{hc}{E}$.For a non-relativistic electron,the energy

Vedclass · Accepted Answer

$\lambda_{Ph} > \lambda_{el}$

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(A) For a photon,the energy $E$ is given by $E = \frac{hc}{\lambda_{Ph}}$,so $\lambda_{Ph} = \frac{hc}{E}$.For a non-relativistic electron,the energy $E$ is given by $E = \frac{p^2}{2m}$,where $p$ is the momentum. The de Broglie wavelength is $\lambda_{el} = \frac{h}{p}$.From the energy equation,$p = \sqrt{2mE}$,so $\lambda_{el} = \frac{h}{\sqrt{2mE}}$.Comparing the two expressions:$\lambda_{Ph} = \frac{hc}{E}$$\lambda_{el} = \frac{h}{\sqrt{2mE}}$Given the same energy $E$,the ratio is $\frac{\lambda_{Ph}}{\lambda_{el}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{c\sqrt{2mE}}{E} = c \sqrt{\frac{2m}{E}}$.Since $c$ is the speed of light $(3 	imes 10^8 \ m/s)$ and $E$ is very small $(10^{-20} \ J)$,the value of $\lambda_{Ph}$ is significantly larger than $\lambda_{el}$.Therefore,$\lambda_{Ph} > \lambda_{el}$.

$A$ photon and an electron are given the same energy $(10^{-20} \ J)$. If the wavelengths associated with the photon and the electron are $\lambda_{Ph}$ and $\lambda_{el}$ respectively,then which of the following statements is correct?

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