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

Answer

A

Explanation

`E prop m " " and " " rprop(1)/(m)`
i.e., energy will becomes two times and radius will become half.