Question

In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $E_{0}$ and radius $r_{0}$ of the ground state will be...... ( $a_{0}$ is the Bohr's ground state radius of actual $H$ -atom)

• A. $E_{0}=-27.2eV,r_{0}=\frac{a_{0}}{2}$
• B. $E_{0}=-27.2eV,r_{0}=2a_{0}$
• C. $E_{0}=-13.6eV,r_{0}=\frac{a_{0}}{2}$
• D. $E_{0}=-13.6eV,r_{0}=a_{0}$

Answer 1

Answer: (A)

From the Bohr's model, we have the relations for energy and radius as
$E_{n}=-\frac{m e^{4}}{8 \epsilon _{0}^{2} n^{2} h^{2}}$ and $r_{n}=\frac{\epsilon _{0} n^{2} h^{2}}{\pi m e^{2}}$
Hence, $E \propto m$ and $r_{n} \propto \frac{1}{m}$
So, if mass is doubled then energy will get doubled and $r$ will reduce to half.
$\therefore E_{0}=-13.6\times 2=-27.2eV$ and $r_{0}=\frac{a_{0}}{2}$