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(C) The de Broglie wavelength of an electron with kinetic energy $K$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.For an electron,$\la

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$1.24 \, keV$

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(C) The de Broglie wavelength of an electron with kinetic energy $K$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.For an electron,$\lambda_e \approx \frac{12.27}{\sqrt{K}} \, \mathring{A}$,where $K$ is in $eV$.Given $K = 1.5 \, eV$,we have $\lambda_e = \frac{12.27}{\sqrt{1.5}} \, \mathring{A}$.The wavelength of a photon is given by $\lambda_p = \frac{hc}{E_p}$.Since $\lambda_p = \lambda_e$,we have $\frac{hc}{E_p} = \frac{12.27}{\sqrt{1.5}} \, \mathring{A}$.Using $hc \approx 12400 \, eV \cdot \mathring{A}$,we get $E_p = \frac{12400 \cdot \sqrt{1.5}}{12.27} \, eV$.$E_p \approx \frac{12400 \cdot 1.2247}{12.27} \approx 1237.5 \, eV \approx 1.24 \, keV$.

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

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