Question

In a hypothetical Bohr’s hydrogen atom the mass of the electron is doubled. The energy \(E_0\) and radius \(r_0\) of the first orbit will be (\(a_0\) is the Bohr radius for the first orbit):

• A. \(E_0=-27.2\,\text{eV},\; r_0=a_0\)
• B. \(E_0=-13.6\,\text{eV},\; r_0=\dfrac{a_0}{2}\)
• C. \(E_0=-27.3\,\text{eV},\; r_0=\dfrac{a_0}{2}\)
• D. \(E_0=-13.6\,\text{eV},\; r_0=a_0\)

Hint:

In Bohr’s model: \[ r_n \propto \frac{1}{\mu}, \qquad E_n \propto -\mu \] Increasing the electron mass decreases orbital radius and increases binding energy.

Answer 1

The Correct Option is C

Step 1: Recall Bohr model dependences. For a hydrogen atom, the Bohr radius and energy depend on the reduced mass \(\mu\) of the electron–proton system. \[ r_n \propto \frac{1}{\mu}, \qquad E_n \propto -\mu \] For ordinary hydrogen: \[ \mu \approx m_e \]
Step 2: Effect of doubling the electron mass. If the electron mass is doubled: \[ m_e' = 2m_e \Rightarrow \mu' = 2\mu \]
Step 3: Find the new radius of the first orbit. Since: \[ r_0 \propto \frac{1}{\mu} \] \[ r_0' = \frac{r_0}{2} = \frac{a_0}{2} \]
Step 4: Find the new ground-state energy. Since: \[ E_0 \propto -\mu \] Original ground-state energy: \[ E_0 = -13.6\,\text{eV} \] With doubled mass: \[ E_0' = -2 \times 13.6 = -27.2\,\text{eV} \approx -27.3\,\text{eV} \]
Hence, \[ \boxed{E_0=-27.3\,\text{eV},\quad r_0=\dfrac{a_0}{2}} \]