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′=λ0√2 B) λ′=√2λ0 C) λ′=λ0 D) λ′=λ02 Answer: Option AExplanation: 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=h√2mE×1√E as h and m are constants Hence, λ0λ′=√E′E=√2EE=√2 ∴ λ′=λ0√2