In a hypotherical Bohr hydrogen, the mass of the electron is doubled. The energy `E_(0)` and the radius `r_(0)` of the first orbit will be (`a_(0)` is the Bohr radius)

Options

A. `E_0 =-27.2 eV, r_0 = a_0`
B. `E_0=-27.2 eV, r_0 = (a_0)/(2)`
C. `E_0 = -13.6 eV , r_0 = a_0`
D. `E_0 = -13.6 eV, r_0 = (a_0)/(2)`

Answer

B

Explanation

For the bohr atom, radius `r prop 1/m` and energy `prop` m.
`therefore` for ground state `E_0 = -13.6 eV`
If m is doubled , `E_0 = -27.2 eV`
and r will be `r/2 " or " a_0/2`