When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy 
expressed in eV and de-Broglie wavelength 
. The maximum kinetic energy of photoelectron liberated from another metal B by photons of energy 4.70 eV is 
. If the de-Broglie wavelength of these photoelectrons is 
, then choose the wrong option.
A
The work function of A is 2.25 eV
B
The work function of B is 4.20 eV
C
D
Text Solution
Verified by ExpertsThe correct answer is:
D
We know that, 
…… ..(i)
…… ..(ii)
From these two equations, we have
…… ..(iii)
de-Broglie wavelength is given by
or 
0.5 eV
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