When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is $T_{B} = (T_{A} - 1 .5) eV$ . If the de-Broglie wavelength of these photoelectrons $λ_{B} = 2 λ_{A}$ , then the work function of metal B is:

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1.5 eV

4eV

2eV

3eV

answer is B.

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Detailed Solution

$\begin{array}{l} R e l a t i o n b e t w e e n D e - B r o g l i e w a v e l e n g t h a n d K . E . i s \\ λ = \frac{h}{\sqrt{2 (K E) m_{e}}} \\ \Rightarrow λ \propto \frac{1}{\sqrt{K E}} \\ \Rightarrow \frac{λ_{A}}{λ_{B}} = \frac{\sqrt{K E_{B}}}{\sqrt{K E_{A}}} \\ \Rightarrow \frac{1}{2} = \sqrt{\frac{T_{A} - 1.5}{T_{A}}} \\ \Rightarrow T_{A} = 2 e V \\ ∴ K E_{B} = 2 - 1.5 = 0.5 e V \\ N o w, f o r m e t a l B w o r k f u n c t i o n i s \\ ϕ_{B} = E_{B} - K E_{B} \\ \Rightarrow ϕ_{B} = 4.5 - 0.5 = 4 e V \end{array}$

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