| The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is |
| \[\operatorname{B}=\frac{m{{e}^{4}}}{8{{n}^{2}}\varepsilon _{0}^{2}{{h}^{2}}}\] (m = electron mass). If one decides |
| to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be |
| \[\operatorname{B}=\frac{m{{e}^{4}}}{8{{n}^{2}}\varepsilon _{0}^{2}{{h}^{2}}}\] (M= proton mass). |
| This last expression is not correct because |
A) n would not be integral.
B) Bohr-quantisation applies only to electron
C) the frame in which the electron is at rest is not inertial.
D) the motion of the proton would not be in circular orbits, even approximately.
Correct Answer: C
Solution :
| Option [c] is correct. |
| Explanation: In a hydrogen atom, electrons revolving around a fixed proton nucleus have some centripetal acceleration. So that, its frame of reference is non-inertial. In the frame of reference, where the electron is at rest, the given expression is not true as it forms the non- inertial frame of reference. |
| As the mass of an electron is negligible as compared to proton, so the centripetal force cannot provide the electrostatic force, |
| \[{{\operatorname{F}}_{p}}=\frac{{{m}_{p}}{{\operatorname{v}}^{2}}}{r}\] |
| So the given expression is not true, as it forms non-inertial frame of reference due to \[{{m}_{e}}\]<<<\[{{m}_{p}}\] or centripetal force on \[{{\operatorname{F}}_{e}}<<<{{\operatorname{F}}_{p}}\] |
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