The given data is
q=1.6×10−19C4πϵ01=9×109Nm2C−2
The formula for the value of energy in a hydrogen atom is given by
E=8πϵ0rq2⇒E=2(4πϵ0)rq2
⇒r=12.8×29×109×1.6×10−19
⇒r=169×10−10m
It is given that r=x9×10−10m⇒169×10−10=x9×10−10⇒x=16
JEE Main 2023 — Physics Electromagnetism
Experimentally it is found that 12.8eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is x9×10−10m.
The value of the x is: _____.
(1eV=1.6×10−19J,4πϵ01=9×109C2Nm2 and electronic charge =1.6×10−19C)
Held on 6 Apr 2023 · Verified 6 Jul 2026.
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