Question

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.)

• A. $E_0=-27.2eV,r_0=\frac{a_0}{2}$
• B. $E_0=-27.2eV,r_0=a_0$
• C. $E_0=-13.6eV,r_0=\frac{a_0}{2}$
• D. $E_0=-13.6eV,r_0=a_0$

Answer 1

Answer: (A)

The radius of $n^{th}$ orbit is given by $\frac{h^2{ϵ}_o{n}^2}{\pi m{e}^2}$ .
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 $={\frac{-2{\pi }^2{K}^2{e}^{4}mZ}{h^2{n}^2}}^2$ .
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, $E_0=-2\times 13.6$
$E_0=-27.2eV$ .