Energy of an electron and a photon are equal. | vedclass

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(A) The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2mE}}$.Given $\lambda_e = 1\,\mathring{A}$,the ki

Vedclass · Accepted Answer

$82.67$

Vedclass · Answer

(A) The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2mE}}$.Given $\lambda_e = 1\,\mathring{A}$,the kinetic energy $E$ of the electron is $E = \frac{h^2}{2m\lambda_e^2}$.Using the relation $E = \frac{150}{\lambda_e^2} 	ext{ eV}$ (where $\lambda_e$ is in $\mathring{A}$),we get $E = 150 	ext{ eV}$.Since the energy of the photon is equal to the energy of the electron,$E_{ph} = 150 	ext{ eV}$.The wavelength of the photon is given by $\lambda_{ph} = \frac{hc}{E_{ph}}$.Using $hc \approx 12400 	ext{ eV}\cdot\mathring{A}$,we have $\lambda_{ph} = \frac{12400}{150} \mathring{A} \approx 82.67 \mathring{A}$.

Energy of an electron and a photon are equal. If the wavelength of the electron is $1\,\mathring{A}$,then the wavelength of the photon is approximately ............. $\mathring{A}$.

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