Question

The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom?
A) He+ (n=2)
B) Li2+ (n=2)
C) Li2+ (n=3)
D) Be3+ (n=2)
where n=orbit

Answer 1

A. He+ is a Helium atom missing one electron giving it the same configuration as Hydrogen.

Answer 2

The question that you need to ask is how does the radial distance depend on a) the nuclear charge and b) the principle quantum number "n" of the orbital. If you will forgive a classical derivation:
\[ F = \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2} \]
and so
\[r = \frac{Ze^2}{4\pi\epsilon_0m_ev^2} \]
Recognizing that the orbital momentum is quantized, we have
\[L = mvr = n\hbar \rightarrow v= \frac{n\hbar}{m_er}\]
and substituting this in,
\[r = \frac{Ze^2m_e^2r^2}{4\pi\epsilon_0m_en^2\hbar^2} \]
and simplifying things a tad, we find
\[r = \frac{4\pi\epsilon_0 n^2 \hbar^2}{Ze^2m_e} \propto \frac{n^2}{Z} \]
So the important thing to take away is that the radius is proportional to n^2 and inversely proportional to Z.

Answer 3

In the ground state of hydrogen, n = 1 and Z = 1, so that's just one -> it follows that if you want any other electron orbital to have that same radius, then the principle quantum number of the orbital occupied by the electron needs to equal the square root of the atomic number. This is true only in Beryllium 3+ (Z = 4, n= 2).

Answer 4

Oh goodness, I meant angular momentum, not orbital momentum... :)