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The light of wavelength $\lambda^{\prime}$ incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light, h=Planck's constant, m=mass of electron $)$
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The correct answer is:
$\left[\frac{2(\mathrm{~h} \mathrm{c}-\lambda \phi)}{\mathfrak{m \lambda}}\right]^{\frac{1}{2}}$
(C)
$\mathrm{hv}-\phi=\mathrm{E}_{\max }=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{\mathrm{hc}}{\lambda}-\phi=\frac{1}{2} \mathrm{mv}^{2}$
$\mathrm{v}^{2}=2 \frac{(\mathrm{hc}-\lambda \phi)}{\lambda \mathrm{m}}$
$\mathrm{hv}-\phi=\mathrm{E}_{\max }=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{\mathrm{hc}}{\lambda}-\phi=\frac{1}{2} \mathrm{mv}^{2}$
$\mathrm{v}^{2}=2 \frac{(\mathrm{hc}-\lambda \phi)}{\lambda \mathrm{m}}$
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