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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 8 of 27
Marks: +1, -0
A photon emitted during the de-excitation of electron from a state nn to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV2\,\text{eV}, in a photo cell, with a stopping potential of 0.55 V0.55\,\text{V}. Obtain the value of the quantum number of the state nn.
OR
A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV12.5\,\text{eV} energy. Find out the maximum number of lines emitted by the atom from its excited state.
Solution:  
● For writing Einstein's photoelectric equation
● For writing, En=−  13.6n2E_n=-\;\frac{13.6}{n^2}
● For finding the value of nn
From photoelectric equation, hv=ϕ0+eVsh v = \phi_0 + e V_s
=2+0.55=2.55 eV=2+0.55=2.55\,\text{eV}
Given, En=  13.6n2E_n=\;\frac{13.6}{n^2}
The energy difference
  ΔE=−3.4−(−2.55) eV\;\Delta E=-3.4-(-2.55)\,\text{eV} =−0.85 eV=-0.85\,\text{eV}
  ∴      −13.6n2=−0.85\;\therefore\;\;\;\frac{-13.6}{n^2}=-0.85
∴ n=4\therefore\ n=4
OR
Calculation of energy in excited state Formula
Finding out the maximum number of lines.
Energy in ground state, E1=−13.6 eVE_1=-13.6\,\text{eV}
Energy supplied =12.5 eV=12.5\,\text{eV}
Energy in excited state, −13.6+12.5=−1.1 eV-13.6+12.5=-1.1\,\text{eV}
But, En=  −13.6n2=−1.1E_n=\;\frac{-13.6}{n^2}=-1.1
n=3n=3
Maximum number of lines =3=3
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