From quantisation of angular momentum one gets for hydrogen atom, the radius of the `n^th` orbit as `r_n=((n^2)/(m_e))((h)/(2pi))^2((4pi^2epsilon_0)/(e^2))`
For a hydrogen like atom of atomic number Z,

Options

A. the radius of the first orbit will be the same
B. `r_n` will be greater for larger Z values
C. `r_n` will be smaller for larger Z values
D. none of these

Answer

C

Explanation

For an atom with a single electron, bohr atom model is applicable.
As the value of attraction between a proton and electron is proportional to `e^2`, for an ion with a single electron, `(e^2)/(4piepsilon_0)` is replaced by `(Ze^2)/(4piepsilon_0)` i.e. `r_(n)prop(n^2)/(Z)`.