Q. In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $E_{0}$ and the radius $r_{0}$ of the first orbit will be ($a_{0}$ is the Bohr radius)
A) $E_{0} = - 27.2\ eV;r_{0} = a_{0}/2$
B) $E_{0} = - 27.2\ eV;r_{0} = a_{0}$
C) $E_{0} = - 13.6\ eV;r_{0} = a_{0}/2$
D) $E_{0} = - 13.6\ eV;r_{0} = a_{0}$
Explanation
(a) Here radius of electron orbit $r \propto 1/m$ and energy $E \propto m$, where $m$ is the mass of the electron
Hence energy of hypothetical atom
$E_{0} = 2 \times ( - 13.6eV) = - 27.2eV$ and radius $r_{0} = \frac{a_{0}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.