Bohr Model Of The Hydrogen Atom

For hydrogen-like species, the radius of orbit is given by: Where:
- = principal quantum number
- = atomic number
- = Bohr radius constant
For hydrogen (H) in ground state: For He (Z = 2) in state: Now, Correction:
We must compare radii in pm.
Let’s refine using:
- H atom: , :
- He : , :
\ So: But image marked correct as , which implies: So final correct interpretation:
For ground state of H vs. state of He , it is:

We need to count electrons with magnetic quantum number , which corresponds to:
- s-orbitals ( ) →
- p-orbitals ( ) →
- d-orbitals ( ) →
⇒ Each orbital has one orbital
Now determine electron configurations:
For Fe (Atomic number 26 → loses 3e⁻ → 23e⁻)
Electronic configuration:
Electrons in orbitals:
- 1s, 2s, 3s → 3 orbitals × 1 = 3 electrons
- 2p, 3p → each has 1 orbital with → 2
- 3d: fills each orbital singly → includes 1 in → 1
Total =
(But this is incorrect with our original assumption—let's reanalyze with all orbitals that have )
Actually, for each full orbital set:
- s: 1 orbital → 2 electrons (all with )
- p: 3 orbitals → 1 has → max 2 electrons
- d: 5 orbitals → 1 with
Let’s sum for each shell:
Fe : 23 electrons
- 1s² → 2 (m_l = 0) → 2
- 2s² → 2 (m_l = 0) → 2
- 2p⁶ → 2 (in )
- 3s² → 2 (m_l = 0) → 2
- 3p⁶ → 2 (in )
- 3d⁵ → 1 (each orbital singly filled) → 1 in
Total =
Cr : Z = 24, loses 1e⁻ → 23e⁻ Expected config: (Same as Fe )
→ So total is also 11