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

E_{o} = - 27.2 e V; r_{o} = a_{o} / 2

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E_{o} = - 27.2 e V; r_{o} = a_{o}

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E_{o} = - 13.6 e V; r_{o} = a_{o} / 2

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E_{o} = - 13.6 e V; r_{o} = a_{o}

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Solution

The correct option is A $E_{o} = - 27.2 e V; r_{o} = a_{o} / 2$
$\begin{matrix} Given-* Mass of the e^{-} (m) is doubled we know that Energy of an e^{-} is given by- E_{0} = - \frac{m e^{4} z^{2}}{8 ε_{0}^{2} n^{2} h^{2}} = - 13.6 ev (at ground state) \end{matrix}$

$\begin{matrix} if mass is doubled then - E_{0} is also double \Rightarrow [E_{f} = - 27.2 e V] we also know that radius of the e^{-} is given by (in an orbit) - orbitat radius re = \frac{n^{2} h^{2} ε_{0}}{π m z e^{2}} = a_{0} (bohr radius n=1) [r = \frac{a_{0}}{2}] If the mass is doubled \end{matrix}$

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