Question

Directions: The following question has four choices, out of which one or more are correct.When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have minimum kinetic energy TA expressed in eV and de Broglie wavelength λA. The maximum kinetic energy ofphotoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA−1.50eV). If the de Broglie wavelength of these photoelectrons is λB=2λA, then

• A. (A) the work function of A is 2.25 eV
• B. (B) the work function of B is 4.20 eV
• C. (C) TA = 2.00 eV
• D. (D) TB = 2.75 eV

Answer 1

Answer: (A), (D)

(A) the work function of A is 2.25 eV

Answer 2

Explanation:
Kmax=E−WTherefore,TA=4.25−WA… (i)TB=(TA−1.50)=4.70−WB… (ii)From Eqs. (i) and (ii),WB−WA=1.95eV… (iii)de Broglie wavelength is given byλ=h2Km or λ∝1KλAλB∝KBKA2=TATA−1.5 This gives TA=2eV From equation (i), WA=4.25−TA=2.25eV From equation (iii), WB=WA+1.95eV=(2.25+1.95)eV Or, WB=4.20eVTB=4.70−WB=4.70−4.20=0.50eV