Work function of a metal X equals the ionization energy of $L i^{2 +}$ ion in second excited state. Work function of another metal Y equals the ionization energy of $H e^{+}$ ion with electron in $n = 4$ . Now photons of energy E fall separately on both the metals such that maximum kinetic energy of photoelectron emitted from metal X is half that of photoelectron emitted from metal Y. Choose the correct statement $(E_{0} =$ Potential energy of electron in ground state of hydrogen atom)

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$E = - 3.5 E_{0}$

If E is increased, difference in maximum kinetic energies of photo electrons emitted from X and Y increases

$E = \frac{- 7 E_{0}}{8}$

If E is increased, difference in maximum kinetic energies of photo electrons emitted from X and Y remains constant

answer is B, D.

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Detailed Solution

Energy of H-atom in ground stale $= \frac{F_{0}}{2}$

$ϕ_{X} = \frac{F_{0}}{2} \{for L_{i}, \begin{matrix} z = 3 \\ n = 3 \end{matrix}\}$

$ϕ_{Y} = (\frac{E_{0}}{2}) \times \frac{4}{16} {for He \begin{matrix} z = 2 \\ n = 4 \end{matrix}}$

$k ε_{X} = E - Q_{X}$

$k ε_{Y} = E - Q_{Y}$

and $K E_{X} = \frac{K E_{Y}}{2}$

Solving we get the answer

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