1)

 In photoelectric  effect, initally when energy of electrons emitted is E0 , de-Broglie wavelength associated with them is λ0 , Now , energy is doubled then associated de-Broglie wavelength λ' is 


A) λ=λ02

B) λ=2λ0

C) λ=λ0

D) λ=λ02

Answer:

Option A

Explanation:

 de-Broglie wavelength is given by

 λ=hp , where h= planck's constant and 

 p= momentum

 also, energy (E) and momentum are related as

 E=p22m

     p=2mE

      λ=h2mE×1E  as h and m are constants

 Hence,  λ0λ=EE=2EE=2

     λ=λ02