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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 18
Total: 37
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
Solution:  
(a) U =
1
4πε0
e×−e
r
 = - 9 × 109 ×
(1.6×10−19)2
0.53×10−10
 J
or U =
−9×109×(1.6×10−19)2
0.53×10−10×1.6×10−19
 eV or U = - 27.2 eV
(b) K.E. = +
|U|
2
 = 13.6 eV
∴ Total energy E = P.E + K.E = –27.2 + 13.6 = –13.6 eV
Now W = ΔU = 0 – (–13.6) or W = +13.6 eV
(c) P.E. at 1.06 × 10–10 m separation,
U' =
9×109×(−1.6×10−19)×(1.6×10−19)
1.06×10−10
= - 21.74 × 10−19 J
or U' =
−21.74×10−19
1.06×10−19
 = - 13.585 eV
For (a): Taking –13.585 eV as zero of P.E., then
P.E. of the system = –27.17 – (–13.585) = –13.585 eV = –13.6 eV.
For (b): DU = – 13.6 eV – (–13.6 eV) = 0
∴ W = 0
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