Discussion Forum



1)

 In photoelectric  effect, initally when energy of electrons emitted is E0 , de-Broglie wavelength associated with them is $\lambda$0 , Now , energy is doubled then associated de-Broglie wavelength $\lambda$' is 


A) $\lambda^{'}=\frac{\lambda_{0}}{\sqrt{2}}$

B) $\lambda^{'}=\sqrt{2}\lambda_{0}$

C) $\lambda^{'}=\lambda_{0}$

D) $\lambda^{'}=\frac{\lambda_{0}}{2}$

Answer:

Option A

Explanation:

 de-Broglie wavelength is given by

 $\lambda=\frac{h}{p}$ , where h= planck's constant and 

 p= momentum

 also, energy (E) and momentum are related as

 $E=\frac{p^{2}}{2m}$

 $\therefore$    $p=\sqrt{2mE}$

 $\therefore$     $\lambda= \frac{h}{\sqrt{2mE}}\times\frac{1}{\sqrt{E}}$  as h and m are constants

 Hence,  $\frac{\lambda_{0}}{\lambda'}=\sqrt{\frac{E'}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$

 $\therefore$    $\lambda'= \frac{\lambda_{0}}{\sqrt{2}}$