In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy E0 prime and radius r0 prime of the first orbit will be ( r 0 is the Bohr radius)

Question

In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy E0 prime and radius r0 prime of the first orbit will be ( r 0 is the Bohr radius) (A) -11.2 eV (B) -6.8 eV (C) -13.6 eV (D) -27.2 eV

Tardigrade Admin · Accepted Answer

As r propto (1/ m ) ∴ r 0 prime=(1/2) r 0 As E propto m ∴ E 0 prime=2(-13.6)=-27.2 eV

Q. In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy $E_{0}^{'}$ and radius $r_{0}^{'}$ of the first orbit will be ( $r_{0}$ is the Bohr radius)

$- 11.2 e V$

$- 6.8 e V$

$- 13.6 e V$

$- 27.2 e V$

As $r \propto \frac{1}{m}$
$∴ r_{0}^{'} = \frac{1}{2} r_{0}$
As $E \propto m$
$∴ E_{0}^{'} = 2 (- 13.6) = - 27.2 e V$

Q. In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy E0′​ and radius r0′​ of the first orbit will be ( r0​ is the Bohr radius)

−11.2eV

−6.8eV

−13.6eV

−27.2eV