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Q.

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy TA and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = TA-1.50 eV. If the de Broglie wavelength of these photoelectrons is λB=2λA, then

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a

the work function of A is 2.25 eV

b

the work function of B is 4.20 eV

c

TA=2.00eV

d

TB=2.75eV

answer is A.

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Detailed Solution

For metal A:4.25=WA+TA ……..(i) Also, TA=12mvA2=12m2vA2m=pA22m=h22mλA2 ………(ii) ∵λ=hp For metal B:4.7=TA−1.5+WB ………(iii) Also, TB=h22mλB2×2mλA2h2=λA2λB2⇒TA−1.5TA=λA22λA2=λA24λA2=14 ∵λB=2λA given ⇒4TA−6=TA⇒TA=2eV∴WA=2.25eV,WB=4.20eV,TB=(2-1.5)eV = 0.5 ev

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