The ratio of the radii of the third Bohr orbit of $He^+$ and the first Bohr orbit of $H$ is
A$3$
B$9$
C$6$
D$9/2$
Answer & Solution
Correct answer: D. $9/2$
$r_n = (0.529\,Å)\,n^2/Z$. For $He^+$ ($Z=2$, $n=3$): $r = 9/2 \cdot 0.529\,Å$. For $H$ ($Z=1$, $n=1$): $r = 0.529\,Å$. Ratio $= 9/2$.
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