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The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals $0.51 \overset{0}{A}$ and its ground state energy equals $- 13.6 e V$ . If the electron in the hydrogen atom is replaced by muon ( $μ^{-}$ ) [charge same as electron and mass 207 $m_{e}$ ], the first Bohr radius and ground state energy will be,

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$25 .6 \times 10^{- 13} m, - 2 .8 eV$

$0 .53 \times 10^{- 13} m, - 3 .6 eV$

$2 .56 \times 10^{- 13} m, - 13 .6 eV$

$2 .56 \times 10^{- 13} m, - 2 .8 keV$

answer is D.

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Detailed Solution

$\begin{array}{l} r_{n} \propto \frac{1}{m} \\ r_{μ} = \frac{0 .51}{207} ≃ 2 .56 \times 10^{- 13} m \\ E \propto m_{e} \\ (E)_{μ} = - 13 .6 \times 207 \\ = - 2 .8 keV \end{array}$

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