The wavelength lambda_e of an electron and la | vedclass

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(A) The de Broglie wavelength of an electron with kinetic energy $E$ is given by:$\lambda_e = \frac{h}{\sqrt{2 m_e E}}$  .... $(i)$The wavelength of a

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$\lambda_p \propto \lambda_e^2$

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(A) The de Broglie wavelength of an electron with kinetic energy $E$ is given by:$\lambda_e = \frac{h}{\sqrt{2 m_e E}}$  .... $(i)$The wavelength of a photon with energy $E$ is given by:$\lambda_p = \frac{hc}{E}$  or  $E = \frac{hc}{\lambda_p}$  .... $(ii)$From equation $(i)$,squaring both sides:$\lambda_e^2 = \frac{h^2}{2 m_e E}$  or  $E = \frac{h^2}{2 m_e \lambda_e^2}$  .... $(iii)$Equating the expressions for $E$ from $(ii)$ and $(iii)$:$\frac{hc}{\lambda_p} = \frac{h^2}{2 m_e \lambda_e^2}$Rearranging for $\lambda_p$:$\lambda_p = \left( \frac{2 m_e c}{h} \right) \lambda_e^2$Since $\frac{2 m_e c}{h}$ is a constant,we have:$\lambda_p \propto \lambda_e^2$

The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by

Similar Questions

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