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A photo sensitive 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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The correct answer is:
$\sqrt{\frac{h \lambda}{2 m c}}$
de-Broglie wavelength
$\lambda_0=\frac{h}{p}=\frac{h}{\sqrt{2 m E_k}}$
And $E_K=h v=\frac{h c}{\lambda}$
$\therefore \quad \lambda_v=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda}}}=\sqrt{\frac{\lambda h}{2 m c}}$
$\lambda_0=\frac{h}{p}=\frac{h}{\sqrt{2 m E_k}}$
And $E_K=h v=\frac{h c}{\lambda}$
$\therefore \quad \lambda_v=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda}}}=\sqrt{\frac{\lambda h}{2 m c}}$
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