Question

If $\lambda _{ο}$ and $\lambda $ be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

• A. $\sqrt{\frac{2 hc}{m} \left(\frac{\left(\lambda \right)_{ο} - \lambda }{\lambda \left( \lambda \right)_{ο}}\right)}$
• B. $\sqrt{\frac{2 hc}{m} \left(\left(\lambda \right)_{ο} - \lambda \right)}$
• C. $\sqrt{\frac{2 h}{m} \left(\left(\lambda \right)_{ο} - \lambda \right)}$
• D. $\sqrt{\frac{2 h}{m} \left(\frac{1}{\left(\lambda \right)_{ο}} - \frac{1}{\lambda }\right)}$

Answer 1

Answer: (A)

Einstein photo Electric Equation
$\text{hν} - \text{hν}_{0} = \frac{1}{2} \text{mV}_{\text{max}}^{2}$
$\text{h} \frac{\text{C}}{\lambda } - \text{h} \frac{\text{C}}{\lambda _{0}} = \frac{1}{2} \text{mV}_{\text{max}}^{2}$
$\frac{2 \text{hC}}{\text{m}} \left(\frac{1}{\lambda } - \frac{1}{\left(\lambda \right)_{0}}\right) = \text{V}_{\text{max}}^{2}$
$\left(\text{V}\right)_{\text{max}} = \sqrt{\frac{2 \text{hC}}{\text{m}} \left(\frac{\left(\lambda \right)_{0} - \lambda }{\lambda \left( \, \, \, \lambda \right)_{0}}\right)}$