When photons of energy 4.25 eV strike the surface of a metal A , the ejected photoelectrons have maximum kinetic energy $T_{A}$ eV and de-Broglie wavelength $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is $T_{B} = (T_{A} - 1.50 e V)$ . If the de-Broglie wavelength of these photoelectrons is $λ_{B} = 2 λ_{A}$ , then,

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$T_{A} = 2.00 e V$

The work function of B is 4.20 eV

$T_{B} = 2.75 e V$

The work function of A is 2.25 eV

answer is A, B, C.

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Detailed Solution

In photoelectric effect, maximum kinetic energy of the ejected photoelectrons is given by $T = h v - ϕ$ .
Thus,
$T_{A} = 4.25 - ϕ_{A}$ , (1)
$T_{B} = 4.70 - ϕ_{B}$ , (2)
Given,
$T_{B} = T_{A} - 1.50$ , (3)
$\frac{λ_{A}}{λ_{B}} = \frac{h / \sqrt{2 m T_{A}}}{h / \sqrt{2 m T_{B}}} = \sqrt{\frac{T_{B}}{T_{A}}} = \frac{1}{2}$ . (4)
Solve the equations (3) and (4) to get $T_{A} = 2 e V$ and $T_{B} = 0.5 e V$ . Substitute $T_{A}$ in equation (1) to get $ϕ_{A} = 2.25 e V$ and substitute $T_{B}$ in equation (2) to get $ϕ_{B} = 4.2 e V$ .