When photons of energy 4.28 eV strike the surface of a metal, the ejected photoelectrons have a maximum kinetic energy EA eV and de-Broglie wavelength $λ_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is EB = (EA −1.50)eV. If The de-Broglie wavelength of these photoelectrons is $λ_B=2λ_A$, then
Answer & explanation
Correct answer: option 3
As, $λ=\frac{h}{\sqrt{2mE}}$; so $\frac{λ_B}{λ_A}=\sqrt{\frac{E_A}{E_B}}$
or $2=\sqrt{\frac{E_A}{E_B}}$ or $E_A=4E_B$
or $\frac{E_A}{4}=E_A-1.5$ or $E_A=2.0eV$
$\phi_A$ = 4.28 − 2.00 = 2.28eV
$\phi_B$ = 4.70 − 0.50 = 4.20eV