Question

In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy ${\mathrm{E}}_0$ ​and the radius $r_0$ of the first orbit will be ($r_0$ is the Bohr radius)-

• A. -6.8 $\mathrm{eV}$
• B. -11.2 $\mathrm{eV}$
• C. -27.2 $\mathrm{eV}$
• D. -33.6 $\mathrm{eV}$

Answer 1

The correct option is C

-27.2 $\mathrm{eV}$

Explanation for correct answer:-

Option (C) -27.2 $\mathrm{eV}$

Step 1: Radius ${\mathrm{r}}_0$

For the Bohr atom, the radius$\left(\mathrm{r}\right)$ is inversely proportional to the mass of the electron$\left(\mathrm{m}\right)$.

$\mathrm{r}\propto \frac{1}{\mathrm{m}}......\left(1\right)$

Here, the mass of the electron is doubled

So, ${\mathrm{r}}_0=\frac{1}{2\mathrm{m}}....\left(2\right)$

Divide equation (2) by (1),

$$\begin{gathered} \frac{r_0}{r}=\frac{{\displaystyle \frac{1}{2m}}}{{\displaystyle \frac{1}{m}}} \\ \frac{r_0}{r}=\frac{1}{2} \\ {r}_0=\frac{1}{2}r \end{gathered}$$

Therefore, when the mass of an electron is doubled then the radius gets halved.

Step 2: Energy ${\mathrm{E}}_0$

Energy$\left(\mathrm{E}\right)$ is proportional to the mass$\left(\mathrm{m}\right)$ of the electron

$\mathrm{E}\propto \mathrm{m}$

The ground state energy of a Hydrogen atom is $-13.6eV$.

Therefore,

$$\begin{gathered} {\mathrm{E}}_0=2\times (-13.6) \\ =-27.2\mathrm{eV} \end{gathered}$$

Hence, the energy of the first orbit will be $-27.2\mathrm{eV}$.