1.

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,

Answer»

the radius of the first orbit will be the same
`r_n` will be greater for larger Z VALUES
`r_n` will be smaller for larger Z values
none of these

Solution :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)`.


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