In a hypothetical Bohr hydrogen atom, if the | vedclass

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$.Fro

Vedclass · Accepted Answer

$-27.2$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$.From this expression, we can see that the energy $E$ is directly proportional to the mass of the electron, $m$ (i.e., $E \propto m$).Given that the mass of the electron is doubled $(m' = 2m)$, the new energy $E'$ will be $E' = 2 	imes E$.The energy of an electron in the first orbit $(n=1)$ of a standard hydrogen atom is $E_1 = -13.6 	ext{ eV}$.Therefore, the new energy is $E' = 2 	imes (-13.6 	ext{ eV}) = -27.2 	ext{ eV}$.

In a hypothetical Bohr hydrogen atom, if the mass of the electron is doubled, then the energy of the electron in the first orbit is: (in $\text{ eV}$)

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