Q. In a hypothetical Bohr hydrogen, the mass of the electron is doubled. What will be the energy $E_{0}$ and the radius $r_{0}$ of the first orbit? ( $a_{0}$ is the Bohr radius.)
NTA AbhyasNTA Abhyas 2022
Solution:
The radius of $n^{th}$ orbit is given by $\frac{h^{2} \epsilon _{o} n^{2}}{\pi m e^{2}}$ .
So, the radius of $n^{th}$ orbit $ \propto \frac{1}{m}$ .
If mass of the electron is doubled, the radius of $n^{th}$ orbit will be half of its previous value,
$r_{0}=\frac{a_{0}}{2}$ .
Total energy $E$ of $n^{th}$ orbit $=\frac{- 2 \pi ^{2} K^{2} e^{4} m Z}{h^{2} n^{2}}^{2}$ .
So, the total energy of $n^{th}$ orbit $ \propto m$ .
So, as the mass of electron is doubled, the total energy is doubled.
The energy of the first orbit of hydrogen atom $=-13.6eV$ .
For the hypothetical atom, energy, $E_{0}=-2\times 13.6$
$E_{0}=-27.2eV$ .
So, the radius of $n^{th}$ orbit $ \propto \frac{1}{m}$ .
If mass of the electron is doubled, the radius of $n^{th}$ orbit will be half of its previous value,
$r_{0}=\frac{a_{0}}{2}$ .
So, the total energy of $n^{th}$ orbit $ \propto m$ .
So, as the mass of electron is doubled, the total energy is doubled.
The energy of the first orbit of hydrogen atom $=-13.6eV$ .
For the hypothetical atom, energy, $E_{0}=-2\times 13.6$
$E_{0}=-27.2eV$ .