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JEE MainPhysics
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Atoms and Nuclei: JEE Main Physics Question with Solution

Experimentally it is found that 12.8 eV12.8\text{ eV} 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 9x×1010 m\frac{9}{\text{x}}\times 10^{-10}\text{ m}. The value of the x\text{x} is ______\_\_\_\_\_\_. (1eV=1.6×1019J,14π0=9×109 Nm2/C2 and electronic charge=1.6×1019J C)(1\text{eV}=1.6\times 10^{-19}\text{J}, \frac{1}{4\pi\in_0}=9\times 10^9\text{ Nm}^2/\text{C}^2\text{ and electronic charge}=1.6\times 10^{-19}\text{J C})
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Source and academic review
Question type
Numerical
Exam relevance
JEE Main · Physics
Concepts assessed
Physics
Academic status
Reviewed by official_key
Source
pyq
Editorial review
9 September 2026

Students also ask

Why is the binding energy ke2/(2r)ke^2/(2r) instead of ke2/rke^2/r?

The binding energy is the work required to remove the electron to infinity, which equals Etotal-E_{\text{total}}. Since total energy E=K+U=12(ke2r)ke2r=ke22rE = K + U = \frac{1}{2}\left(\frac{ke^2}{r}\right) - \frac{ke^2}{r} = -\frac{ke^2}{2r}, the separation energy is +ke22r+ \frac{ke^2}{2r}.