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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

TB = 2.75 eV

b

the work function of B is 4.20 eV

c

TA = 2.00 eV

d

the work function of A is 2.25 eV

answer is A, B, C.

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Detailed Solution

For metal A, 

4.25=WA+0.TA TA=12m2VA2m=PA22m=h22mλA2

For metal B,  

4.7=TA1.5+WB TB=h22mλB2×2mλA2h2

=λAλB2=λA2λA2=144TA6=TA TA=2eVWA=2.25eVWB=4.20eV TB=3eV

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