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)//2`
B. `E_(0)=- 27.2 eV, r_(0) = a_(0)`
C. `E_(0)=- 13.6 eV, r_(0) = a_(0)//2`
D. `E_(0)=- 13.6 eV, r_(0) = a_(0)`

Answer

A

Explanation

Here radius of electron orbit `r prop 1//m` and energy `E prop m`, where `m` is the mass of the electron.
Hence energy of hypothetical atom
`E_(0) = 2 xx (-13.6eV)=- 27.2 eV` and radius `r_(0) = (a_(0))/(2)`