Discussion Forum



1)

The light of wavelength $\lambda$  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]


A) $\left[\frac{2(hv-\phi)\lambda}{mc}\right]$

B) $\left[\frac{2(hc-\lambda\phi)}{m\lambda}\right]^{1/2}$

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

D) $\left[\frac{2(hc-\phi)}{m\lambda}\right]^{}$

Answer:

Option B

Explanation:

 In photoelectric edffect, the maximum kinetic energy possessed  by the particle.

                  $KE_{max}=hv-\phi$

$\frac{1}{2}mv^{2}=\frac{hc}{\lambda}-\phi$           $[\because v=\frac{c}{\lambda}]$

$\Rightarrow $      $ v^{2}=\frac{2(hc-\phi\lambda)}{\lambda m}$

$  v=\left[\frac{2(hc-\phi\lambda)}{\lambda m}\right]^{1/2}$