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 - Franck-Hertz Experiment

The Franck-Hertz experiment and related scattering experiments show that

1. electrons are always scattered elastically from atoms
2. electrons are never scattered elastically from atoms
3. electrons of a certain energy range can be scattered inelastically, and the energy lost by electrons is discrete
4. electron always lose the same energy when they are scattered inelastically
5. there is no energy range in which the energy lost by electrons varies continuously

(GR8677 #47)

Solution:

Franck–Hertz experiment: confirmed Bohr's quantized model of the atom by demonstrating that atoms could indeed only absorb (and be excited by) specific amounts of energy (quanta).

Answer: C

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

Quantum Mechanics - Bohr Radius

The solution to the Schrodinger equation for the ground state of hydrogen is



where a₀ is the Bohr radius and r is the distance from the origin. Which of the following is the most probable value for r?

A. 0
B. a₀ / 2
C. a₀
D. 2a₀
E. ∞

(GR0177 #93)

Solution:

Probability



The most probable value of r corresponds to the peak of the plot of P(r) versus r.
The slope of the curve at this point is zero.





For  

r = a₀

Answer: C

Nuclear & Particle Physics - Positronium

The positronium “atom” consists of an electron and a positron bound together by their mutual Coulomb attraction and moving about their center mass, which is located halfway between them. Thus the positronium “atom” is somewhat analogous to a hydrogen atom. The ground-state binding energy of hydrogen is 13.6 electron volts. What is the ground-state binding energy of positronium. 

A. (½)²× 13.6 eV 
B. ½ × 13.6 eV 
C. 13.6 eV 
D. 2 × 13.6 eV 
E. (2)²× 13.6 eV 

 (GR8677 #99)

Solution

Hydrogen: proton and electron
Positronium: positron (anti electron) and electron

By definition of particles and antiparticles, the mass of the electron and the positron are the same.
→ The reduced mass Positronium:



→ Energy levels of Positronium is half those of Hydrogen

E_n(H)  = −13.6 / n²
E_n(Ps) = ½ E_n(H) 

For the ground-state n = 1 → E_(Ps) =  − ½ × 13.6 eV 

Answer: B