Practice portal › Atoms › Bohr Model of the Hydrogen Atom

In which of the following system will the radius of the second orbit be minimum?

Asked in RS Academy GUJCET booklet · Bohr's rules in other systems

Answer: (2) Mg¹¹⁺

Step-by-step solution

Idea: for a one-electron (hydrogen-like) system the Bohr radius of the nth orbit is rₙ=(n²)/Za₀, with a₀=0.53 A.

The orbit is fixed at n=2, so r₂=(4a₀)/Z: the larger the nuclear charge, the smaller the orbit.

Mg¹¹⁺ is magnesium (Z=12) with 11 of its 12 electrons removed, so it is hydrogen-like and has the largest Z on the list.

r₂=(4×0.53)/(12)=0.18 A, against 2.12 A for hydrogen and 1.06 A for He⁺.

So the second orbit is smallest in Mg¹¹⁺.

Why the other options are wrong

More Bohr Model of the Hydrogen Atom questionsAll Bohr Model of the Hydrogen Atom questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer