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Question

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV and deBroglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is TB=(TA1.50)eV. If the de Broglie wavelength of these photoelectrons is λB=2λA, then

A
The work function of A is 2.25 eV
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B
The work function of b is 4.20 eV
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C
TA=2.00 eV
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D
TB=2.75 eV
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Solution

The correct options are
A The work function of b is 4.20 eV
B TA=2.00 eV
D The work function of A is 2.25 eV
As we know,
Kmax=EW0
TA=4.25(W0)A
TB=(TA1.5)=4.70(W0)B ...(ii)
Equation (i) and (ii) gives (W0)B(W0)A=1.95 eV
De Broglie wave length λ=h2mKλ1K
λBλA=KAKB2=TATA1.5TA=2 e V
From equation (i) and (iii)
WA=2.25 eV and WB=4.20 eV

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