Quantum Mechanics - Ionization Potential

Which of the following atoms has the lowest ionization potential?

A. ²_₄He
B. ⁷₁₄N
C. ⁸₁₆O
D. ¹⁸₄₀Ar
E. ⁵⁵₁₃₃Cs

(GR9677 #39)

Solution:

Ionization potential: the energy required to to remove the valence (outermost) electron from an atom

It is minimum for atoms (alkali metals) which have weakly bound electrons and higher for the noble gases which have closed shells.

²_₄He and Argon ⁸₄₀Ar are noble gases which have a closed-shell valence electron configuration.

⁷₁₄N → 1s² 2s² 2p⁶ 3s² 3p² (2 outermost electrons, almost a full orbital)

⁸₁₆O → 1s² 2s² 2p⁶⁴ 3s² 3p² (4 outermost electrons, almost a full orbital)

⁵⁵₁₃₃Cs has the highest atomic number and hence a lot of electrons. The valence electron is far from the nucleus (weakly bound), thus it needs the lowest ionization potential to remove it.

Answer: E

Nuclear & Particle Physics - Helium

If a singly ionized Helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level E_f  of the atom, and the n value, of n_f  this final state?

	Ef  (eV)	nf
A.	− 6.0	3
B.	− 6.0	2
C.	− 14	2
D.	− 14	1
E.	− 52	1

(GR9677 #40)

Solution:

E_photon = E_i  − E_f 

Helium: 2 electrons, 2 protons, 2 neutron
Singly ionized Helium, He⁺:1 electrons, 2 protons, 2 neutron
He⁺ → Hydrogen-like atom

Bohr's Equation for Hydrogen-like atom: E_n =  −13.6 Z²/n² eV

For Helium, Z = 2,
E_i = E_(n =4)  = − 13.6 (2)² /4² = − 13.6 /4 ≈ − 3.4 eV

E_photon  = hν = hc / λ 
with
h = 6.63 × 10⁻³⁴ Joule.second = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ  = 470 nm = 470 × 10⁻⁹ m = 4.7 × 10⁻⁷ m

E_photon = (4.1 × 10⁻¹⁵)(3 × 10⁸) / (4.7 × 10⁻⁷) ≈ 3 eV

E_f  = E_i − E_photon  
=  − 3.4 − 3
= − 6.4  eV

To find n_f :

n_f ² = −13.6 Z²/ E_f 
= −13.6 (2)²/ (− 6.4)
= 54.4/6.4 ≈ 9
n_f  = 3

Answer: A

Nuclear & Particle Physics - Helium

The energy required to remove both electrons from the Helium atom in its ground state is 79.0 eV. How much energy is required to ionize Helium (i.e. to remove one electron)?

A. 24.6 eV
B. 39.5 eV
C. 51.8 eV
D. 54.4 eV
E. 65.4 eV

(GR0177 #18)

Solution:

Helium: 2 Protons, 2 Neutrons, 2 Electrons.

The energy required to remove one electron from He in its ground state, leaving behind He⁺ (a Hydrogen like atom)

E_n = 13.6 Z²/n² eV

Z = Helium atomic number = 2
n = 1 (ground state)

E₁ = 13.6(4) eV = 54.4 eV

The energy required to remove both electrons from He in its ground state leaving behind He⁺⁺ ion = 79 eV.

Thus, the energy required to remove one electron: 79 − 54.4 = 24.6 eV

Answer: A