Nuclear & Particle Physics - Positronium

The ground-state energy of positronium is most nearly equal to

A. − 27.2 eV
B. − 13.6 eV
C. − 6.8 eV
D. − 3.4 eV
E. 13.6 eV

(GR9677 #12)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

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 = − 6.8 eV

Answer: C

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 - Positronium

Positronium is the bound state of an electron and a positron. Consider only the states of zero orbital angular momentum l = 0. The most probable decay product of any such state of positronium with spin zero (singlet is)

A. 0 photons
B. 1 photons
C. 2 photons
D. 3 photons
E. 4 photons

(GR9677 #53)

Solution:

The singlet state, s = 0 of Positronium is known as para-Positronium, decays preferentially into two gamma rays.

It can decay into any even number of photons (2, 4, 6, ...), but the probability quickly decreases with the number.

Thus, the most probable decay product of singlet state of positronium is 2 photons.

Answer: C

Nuclear & Particle Physics - Positronium

Given that the binding energy of the hydrogen atom ground state is E₀ = 13.6 eV, the binding energy of n = 2 state of positronium (positron-electron system) is

A. 8E₀
B. 4E₀
C. E₀
D. E₀/4
E. E₀/8

(GR9277 #30)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

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

For n = 2,

E_2(Ps) = ½ × ( − 13.6 / 2²)  = − E₀/8

Answer: E

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

Nuclear & Particle Physics - Positronium

Positronium is an atom formed by an electron and a positron (anti-electron). It is similar to the hydrogen atom, with the positron replacing the proton. If a positronium atom makes a transition from a state with n = 3 to a state with n = 1, the energy of the photon emitted in this transition is closest to

A. 6.0 eV
B. 6.8 eV
C. 12.2 eV
D. 13.6 eV
E. 24.2 eV

(GR0177 #31)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

E_n(H)  = −13.6/n²
E_n(Ps) = ½ E_n(H)  = −6.8/n²

For transition n = 3 to n = 1

E_Ps = −6.8 (1/3² − 1/1²) = (−6.8)(−8/9) = 6.04 eV

Answer: A

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