Question

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is :

(1) 3eV

(2) 2eV

(3) 4eV

(4) 1.5eV

Answer 1

Answer is (3) 4eV

λB = 2λA

TA = 4TB ....(i)

and TB = (TA – 1.5) eV ....(ii)

from (i) and (ii)

3TB 1.5 eV

TB = 0.5 eV

TB = 0.5 eV = 4.5 eV – ϕB

ϕ = 4eV