For a hydrogen-like atom of nuclear charge $Z$, the radius of the $n$th Bohr orbit scales as
A$r_n \propto nZ$
B$r_n \propto n^2/Z$
C$r_n \propto Z/n^2$
D$r_n \propto n/Z^2$
Answer & Solution
Correct answer: B. $r_n \propto n^2/Z$
$r_n = (0.529\,Å)\,n^2/Z$. So radius grows as $n^2$ but shrinks as $Z$ — explaining why He$^+$ has a smaller $n=1$ orbit than H.
Related questions
Calcium-40 (Z=20, A=40) and Argon-40 (Z=18, A=40) are best described as:Atoms of the SAME element with DIFFERENT mass numbers are called:An atom of chlorine has atomic number Z = 17 and mass number A = 37. The number of neutronThe atomic number (Z) of an element is equal to the:The neutron was discovered by:Bohr's model of the atom fixed Rutherford's instability problem by proposing that electronRutherford's gold foil experiment concluded that most of the atom is:The Rydberg constant for hydrogen, in SI units, is