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

- 11.2 e V

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- 6.8 e V

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- 13.6 e V

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- 27.2 e V

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Solution

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

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

The correct option is D −27.2eVAs r∝1m ∴r10=12r0As E∝m ∴E10=2(−13.6)=−27.2 eV

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

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