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

`r_(n) prop (1)/(m), E_(n) prop m`