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

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(A) For a photon,the energy $E$ is related to wavelength $\lambda_{ph}$ by $E = \frac{hc}{\lambda_{ph}}$,so $\lambda_{ph} = \frac{hc}{E}$.For an elect

Vedclass · Accepted Answer

$\lambda_{ph} > \lambda_{el}$

Vedclass · Answer

(A) For a photon,the energy $E$ is related to wavelength $\lambda_{ph}$ by $E = \frac{hc}{\lambda_{ph}}$,so $\lambda_{ph} = \frac{hc}{E}$.For an electron,the de Broglie wavelength $\lambda_{el}$ is given by $\lambda_{el} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Comparing the two expressions:$\lambda_{ph} = \frac{hc}{E}$$\lambda_{el} = \frac{h}{\sqrt{2mE}}$Taking the ratio: $\frac{\lambda_{ph}}{\lambda_{el}} = \frac{hc}{E} 	imes \frac{\sqrt{2mE}}{h} = c \sqrt{\frac{2m}{E}}$.Since $c$ is very large $(3 	imes 10^8 \ m/s)$,the value of $\frac{\lambda_{ph}}{\lambda_{el}}$ is much greater than $1$.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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