In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E0 and radius r0 of the ground state will be...... ( a0 is the Bohr's ground state radius of actual H -atom)

Question

In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E0 and radius r0 of the ground state will be...... ( a0 is the Bohr's ground state radius of actual H -atom) (A) E0=-27.2eV,r0=(a0/2) (B) E0=-27.2eV,r0=2a0 (C) E0=-13.6eV,r0=(a0/2) (D) E0=-13.6eV,r0=a0

Tardigrade Admin · Accepted Answer

From the Bohr's model, we have the relations for energy and radius as

E_{n} = - \frac{m e ^{4}}{8 ϵ _{0}^{2} n ^{2} h ^{2}}

and

r_{n} = \frac{ϵ _{0} n ^{2} h ^{2}}{πm 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.

∴ E_{0} = - 13.6 \times 2 = - 27.2 e V

and

r_{0} = \frac{a _{0}}{2}

Q. 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)

$E_{0} = - 27.2 e V, r_{0} = \frac{a _{0}}{2}$

$E_{0} = - 27.2 e V, r_{0} = 2 a_{0}$

$E_{0} = - 13.6 e V, r_{0} = \frac{a _{0}}{2}$

$E_{0} = - 13.6 e V, r_{0} = a_{0}$

Q. In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E0​ and radius r0​ of the ground state will be...... ( a0​ is the Bohr's ground state radius of actual H -atom)

E0​=−27.2eV,r0​=2a0​​

E0​=−27.2eV,r0​=2a0​

E0​=−13.6eV,r0​=2a0​​

E0​=−13.6eV,r0​=a0​

From the Bohr's model, we have the relations for energy and radius as En​=−8ϵ02​n2h2me4​ and rn​=πme2ϵ0​n2h2​ Hence, E∝m and rn​∝m1​ So, if mass is doubled then energy will get doubled and r will reduce to half. ∴E0​=−13.6×2=−27.2eV and r0​=2a0​​

Q. 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)

$E_{0} = - 27.2 e V, r_{0} = \frac{a _{0}}{2}$

$E_{0} = - 27.2 e V, r_{0} = 2 a_{0}$

$E_{0} = - 13.6 e V, r_{0} = \frac{a _{0}}{2}$

$E_{0} = - 13.6 e V, r_{0} = a_{0}$