Question

The maximum velocity of an electron emitted by light of wavelength X incident on the surface of a metal of work function $ \theta $ , is :

• A. $ {{\left[ \frac{2(h\lambda -\theta )}{m} \right]}^{{1}/{2}\;}} $
• B. $ {{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{{1}/{2}\;}} $
• C. $ \left[ \frac{2(hc-\lambda \phi )}{m} \right] $
• D. $ {{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{{1}/{2}\;}} $

Answer 1

Answer: (B)

From Einsteins photo electric equation. Energy of photon $ =\phi +\frac{1}{2}m{{\upsilon }^{2}} $ $ \frac{hc}{\lambda }=\phi +\frac{1}{2}m{{\upsilon }^{2}} $ $ \frac{1}{2}m{{\upsilon }^{2}}=\left( \frac{hc}{\lambda }=\phi \right) $ $ {{\upsilon }^{2}}=2\left( \frac{hc-\lambda \phi }{m\lambda } \right) $ $ \upsilon ={{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{1/2}} $