If the de Broglie wavelengths of an electron | vedclass

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Given that $\lambda_e = \lamb

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Greater than the kinetic energy of the proton

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Given that $\lambda_e = \lambda_p$,it implies that the momentum of the electron and the proton must be equal: $p_e = p_p = p$.The kinetic energy $(K)$ is related to momentum $(p)$ by the formula $K = \frac{p^2}{2m}$.For the electron: $K_e = \frac{p^2}{2m_e}$.For the proton: $K_p = \frac{p^2}{2m_p}$.Since the mass of the proton $(m_p)$ is much greater than the mass of the electron $(m_e)$,i.e.,$m_p > m_e$,it follows that $\frac{1}{m_e} > \frac{1}{m_p}$.Therefore,$K_e > K_p$. The kinetic energy of the electron is greater than the kinetic energy of the proton.

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

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