The de Broglie wavelengths for an electron an | vedclass

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Question

(A) For an electron with kinetic energy $K$,the de Broglie wavelength is given by $\lambda_{e} = \frac{h}{\sqrt{2mK}}$.From this,we can express the ki

Vedclass · Accepted Answer

$\lambda_{p} \propto \lambda_{e}^{2}$

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(A) For an electron with kinetic energy $K$,the de Broglie wavelength is given by $\lambda_{e} = \frac{h}{\sqrt{2mK}}$.From this,we can express the kinetic energy as $K = \frac{h^{2}}{2m\lambda_{e}^{2}}$.For a photon,the energy is given by $E = K = \frac{hc}{\lambda_{p}}$.Equating the kinetic energies of the electron and the photon: $\frac{h^{2}}{2m\lambda_{e}^{2}} = \frac{hc}{\lambda_{p}}$.Rearranging for $\lambda_{p}$,we get $\lambda_{p} = \frac{2mc}{h} \lambda_{e}^{2}$.Since $2, m, c,$ and $h$ are constants,we have $\lambda_{p} \propto \lambda_{e}^{2}$.

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?

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