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. ${\left[\frac{2(hc-\lambda \varphi )}{m\lambda }\right]}^{1/2}$
• C. $\left[\frac{2(hc-\lambda \varphi )}{m}\right]$
• D. ${\left[\frac{2(hc-\lambda \varphi )}{m\lambda }\right]}^{1/2}$

Answer 1

Answer: (B)

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