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(B) The de-Broglie wavelength $\lambda_e$ of an electron is given by $\lambda_e = \frac{h}{p}$,where $p$ is the momentum of the electron.Since the kin

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$\sqrt{\frac{h \lambda}{2 m c}}$

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(B) The de-Broglie wavelength $\lambda_e$ of an electron is given by $\lambda_e = \frac{h}{p}$,where $p$ is the momentum of the electron.Since the kinetic energy $E_k$ of the emitted electron is related to momentum by $E_k = \frac{p^2}{2m}$,we have $p = \sqrt{2m E_k}$.Thus,$\lambda_e = \frac{h}{\sqrt{2m E_k}}$.Given that the work function is neglected,the entire energy of the incident photon is converted into the kinetic energy of the electron: $E_k = E_{photon} = \frac{hc}{\lambda}$.Substituting $E_k$ into the de-Broglie wavelength formula:$\lambda_e = \frac{h}{\sqrt{2m \left(\frac{hc}{\lambda}\right)}}$.Simplifying the expression:$\lambda_e = \sqrt{\frac{h^2}{2m \frac{hc}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h \lambda}{2mc}}$.

$A$ photosensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface,$m$ is mass of the electron,$h$ is Planck's constant,$c$ is the velocity of light).

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