The wavelength of a photon is equal to the wa | vedclass

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(C) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.For a photon,t

Vedclass · Accepted Answer

$1.24\, keV$

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(C) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.For a photon,the wavelength is given by $\lambda = \frac{hc}{E_{ph}}$,where $E_{ph}$ is the energy of the photon.Given that the wavelengths are equal,we have $\frac{h}{\sqrt{2mK}} = \frac{hc}{E_{ph}}$.Squaring both sides,we get $\frac{h^2}{2mK} = \frac{h^2c^2}{E_{ph}^2}$.Rearranging for $E_{ph}^2$,we get $E_{ph}^2 = 2mKc^2$.Substituting the values: $m = 9.1 	imes 10^{-31}\, kg$,$c = 3 	imes 10^8\, m/s$,and $K = 1.5\, eV = 1.5 	imes 1.6 	imes 10^{-19}\, J$.$E_{ph}^2 = 2 	imes (9.1 	imes 10^{-31}) 	imes (1.5 	imes 1.6 	imes 10^{-19}) 	imes (3 	imes 10^8)^2$.$E_{ph}^2 = 2 	imes 9.1 	imes 1.5 	imes 1.6 	imes 9 	imes 10^{-31-19+16} = 393.12 	imes 10^{-34} = 3.9312 	imes 10^{-32}\, J^2$.$E_{ph} = \sqrt{3.9312 	imes 10^{-32}} \approx 1.98 	imes 10^{-16}\, J$.Converting to $eV$: $E_{ph} = \frac{1.98 	imes 10^{-16}}{1.6 	imes 10^{-19}} \approx 1237.5\, eV \approx 1.24\, keV$.

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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