Question

From quantisation of angular momentum, one gets for hydrogen atom, the radius of the $n^{th}$ orbit as$r_{n} = \left( \frac{n^{2}}{m_{e}} \right)\left( \frac{h}{2\pi} \right)^{2}\left( \frac{4\pi^{2}\varepsilon_{0}}{e^{2}} \right)$For a hydrogen like atom of atomic number Z,

• 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 1

Answer: (C)

$r_{n}$will be smaller for larger Z values

Answer 2

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, $\frac{e^{2}}{4\pi\varepsilon_{0}}$ is replaced by $\frac{Ze^{2}}{4\pi\varepsilon_{0}}$

i.e. $r_{n} \propto \frac{n^{2}}{Z}$