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Question

When protons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA expressed in eV 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 isλB=2λA, then

A
Work function of A is 2.25 eV
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B
Work function of B is 4.20 eV
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C
TA = 2.0 eV
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D
TB = 2.75 eV
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Solution

The correct options are
A Work function of A is 2.25 eV
B Work function of B is 4.20 eV
C TA = 2.0 eV
Kmax=EWTA=4.25WAGiven TB=TA1.50=4.70WBWBWA=1.15eVλ1K,λBλA=KAKB2=TATA1.5TA=2eVWA=2.25eV,WB=4.20eV,TB=0.50eV

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