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 - Spectroscopic Notation

In a ³S state of the helium atom, the possible values of the total electronic angular momentum quantum number are

A. 0 only
B. 1 only
C. 0 and 1 only
D. 0, 1/2, and 1
E. 0, 1, and 2

(GR9277 #31)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

• N = the principal quantum number and will often be omitted
• s = the total spin quantum number
• L = the orbital angular momentum quantum number, l but is written as S, P, D, F, … for l = 0, 1, 2, 3,⋯
• j = the total angular momentum quantum number

For ³S 
S → l = 0
3 → 3 = 2s + 1 → s = 1
j = l + s = 0 + 1 = 1

Answer: B 

Nuclear & Particle Physics - Spin

The ground state of the helium atom is a spin

A. singlet
B. doublet
C. triplet
D. quartet
E. quintuplet

(GR9277 #59)

Solution:

Spectroscopic notation: N^2s+1 L_j

2s + 1 = multiplicity

Singlet: 2s + 1 = 1, s = 0
Doublet: 2s + 1 = 2, s = 1/2
Triplet: 2s + 1 = 3, s = 1

Electron configuration of Helium: 1s²

Pauli Exclusion Principle: no two electrons can have exactly the same quantum number.

m_s1 =  ½  and m_s2 =  −½
Total spin quantum number, s = ½ + (−½) = 0
Multiplicity, 2 · 0 + 1 = 1 → Singlet

Answer: A 

Thermal Physics - Probability

A sample of N atoms of helium gas is confined in a 1.0 cubic meter volume. The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container is

A. 0
B. (10⁻⁶)^N
C. (1 − 10⁻⁶)^N
D. 1 − (10⁻⁶)^N
E. 1

(GR8677 #15)

Solution:

Total Probability: P = P₁ + P₂ = 1
P₁ = The probability that one atom is in a 10⁻⁶ cubic meter volume of the container
P₂ =  The probability that none of the helium atoms is in a 10⁻⁶ cubic meter volume of the container
→ P₂ = 1 − P₁

P₁ = 1/n
n = number of 10⁻⁶ m³ cubes in the 1 m³ volume
10⁻⁶ × n = 1 → n = 1/10⁻⁶ = 10⁶ cubes
→ P₁ = 1/n = 1/10⁶ = 10⁻⁶
→ P₂ = 1 − P₁ = 1 − 10⁻⁶
For N atoms: P₂ = (1 − 10⁻⁶)^N

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

Nuclear & Particle Physics - The Sun’s Energy

The primary source of the sun’s energy is a series of thermonuclear reactions in which the energy produces is c² times the mass difference between

A. two hydrogen atoms and one helium atom
B. four hydrogen atoms and one helium atom
C. six hydrogen atoms and two helium atoms
D. three helium atoms and one carbon atom
E. two hydrogen atoms plus two helium atoms and one carbon atom

(GR0177 #19)

Solution:

In the Sun’s core, the protons of the hydrogen atoms (which largely make up the Sun) collide into each other and stick together or “fuse” to create helium nuclei.

4H → He + energy

It takes 4H nuclei to create a He nucleus due to the conservation of atomic mass.

Hydrogen atomic mass = 1 proton = 1.
Helium atomic mass = 2 protons + 2 neutrons = 4.

Answer: B