If λ∘ & beλ threshold wavelength & wavelength of incident light, the velocity of photo electron ejected from the metal Surface is

# If ${\mathrm{\lambda }}_{\circ }$ & be$\mathrm{\lambda }$ threshold wavelength & wavelength of incident light, the velocity of photo electron ejected from the metal Surface is

1. A

$\sqrt{\frac{2\mathrm{h}}{\mathrm{m}}\left({\mathrm{\lambda }}_{\circ }-\mathrm{\lambda }\right)}$

2. B

$\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left({\mathrm{\lambda }}_{\circ }-\mathrm{\lambda }\right)}$

3. C

$\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_{\circ }-\mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_{\circ }}\right)}$

4. D

$\sqrt{\frac{2\mathrm{h}}{\mathrm{m}}\left(\frac{1}{{\mathrm{\lambda }}_{\circ }}-\frac{1}{\mathrm{\lambda }}\right)}$

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### Solution:

48. K.E of the ejected electron is given by equation $\mathrm{h\upsilon }={\mathrm{h\upsilon }}_{0}+\frac{1}{2}{\mathrm{mv}}^{2}$
$\frac{1}{2}{\mathrm{mv}}^{2}=\frac{\mathrm{hc}}{\mathrm{\lambda }}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{0}}=\mathrm{hc}\left(\frac{{\mathrm{\lambda }}_{0}-\mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_{0}}\right)$ $\mathrm{v}=\sqrt{\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_{0}-\mathrm{\lambda }}{{\mathrm{\lambda }}_{0}\mathrm{\lambda }}\right)}$