Work function of metal A is equal to the ionization energy of hydrogen atom in the first excited state. Work function of metal B is equal to the ionization energy of $H e^{+}$ ion in the second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectron emitted from B.
Find the value of E(in eV)

20.8

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32.2

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24.6

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23.8

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Solution

The correct option is D 23.8
$ϕ_{A} = (I . E)_{A}$ = Ionisation energy of electron in $2^{n d}$ orbit of hydrogen atom (first excited state)
$= 3.4 e V [u s e : Δ E = 13.6 [\frac{1}{n^{2}} - \frac{1}{\infty}] e V]$
$ϕ_{B} = (I . E)_{B}$ = Ionisation energy of electron in $2^{n d}$ orbit of $H e^{+}$ ion
=13.6 eV
Given, $K_{A} = 2 K_{B}$
or, $E - ϕ_{A} = 2 (E - ϕ_{B})$
or, $E = 2 ϕ_{B} - ϕ_{A}$
=23.8 eV

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Work function of metal A is equal to the ionization energy of hydrogen atom in the first excited state. Work function of metal B is equal to the ionization energy of $H e^{+}$ ion in the second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectron emitted from B.
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