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In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E₀ and radius r₀ of the first orbit will be (a₀ is the Bohr radius)

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$E_{0} = - 27.2 eV, r_{0} = a_{0} / 2$

$E_{0} = - 27.2 eV, r_{0} = a_{0}$

$E_{0} = - 13.6 eV r_{0} = a_{0} / 2$

$E_{0} = - 13.6 eV, r_{0} = a_{0}$

answer is A.

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Detailed Solution

$E \propto m and r \propto \frac{1}{m}$

i.e. energy will become two times and radius will become half.
Hence, $E_{0} = 2 (- 13.6) = - 27.2 eV and r_{0} = a_{0} / 2$

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