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

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the work function of A is 2.25 eV

the work function of B is 4.20 eV

$T_{A} = 2 .00 eV$

$T_{B} = 2 .75 eV$

answer is A.

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Detailed Solution

$For metal A : 4 .25 = W_{A} + T_{A}$ ……..(i)

$Also, T_{A} = \frac{1}{2} {mv}_{A}^{2} = \frac{1}{2} \frac{m^{2} v_{A}^{2}}{m} = \frac{p_{A}^{2}}{2 m} = \frac{h^{2}}{2 {mλ}_{A}^{2}}$ ………(ii)

$[∵ λ = \frac{h}{p}]$

$For metal B : 4 .7 = (T_{A} - 1 .5) + W_{B}$ ………(iii)

$Also, T_{B} = \frac{h^{2}}{2 {mλ}_{B}^{2}} \times \frac{2 {mλ}_{A}^{2}}{h^{2}} = \frac{λ_{A}^{2}}{λ_{B}^{2}}$

$\begin{array}{l} \Rightarrow \frac{T_{A} - 1 .5}{T_{A}} = \frac{λ_{A}^{2}}{{(2 λ_{A})}^{2}} = \frac{λ_{A}^{2}}{4 λ_{A}^{2}} = \frac{1}{4} [∵ λ_{B} = 2 λ_{A} given] \\ \Rightarrow 4 T_{A} - 6 = T_{A} \Rightarrow T_{A} = 2 eV \\ ∴ W_{A} = 2 .25 eV, W_{B} = 4 .20 eV, T_{B} = (2 - 1.5) eV = 0.5 e v \end{array}$