UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
PHY3702 ASSIGNMENT
Modern Physics — Blackbody Radiation, Photoelectric Effect, Compton Scattering,
de Broglie Wavelength, Linear Algebra and Quantum Mechanics
⋄
Module: Quantum Physics
Moodule Code: PHY3702
Assignment No.: 1
Due Date: May 2026
Submitted in partial fulfilment of the requirements for Modern Physics
at the University of South Africa.
,UNISA | Quantum Physics PHY3702 Assignment
Question 1: Blackbody Radiation — Stefan-Boltzmann Law and Wien’s Displace-
ment Law
Question: Assuming that a given star radiates like a blackbody, estimate:
1. the temperature at its surface,
2. the wavelength of its strongest radiation,
when it emits a total intensity of I = 575 MW m−2 .
1.1 Given Information
I = 575 MW m−2 = 575 × 106 W m−2
The Stefan-Boltzmann law governs the total power radiated per unit area by a blackbody:
I = σT 4
where the Stefan-Boltzmann constant is:
σ = 5.67 × 10−8 W m−2 K−4
1.2(a) Surface Temperature
Rearrange the Stefan-Boltzmann law to solve for T 4 :
I
T4 =
σ
Substitute the known values:
575 × 106
T4 =
5.67 × 10−8
5.75 × 108
T4 =
5.67 × 10−8
Page 1 of 35
,UNISA | Quantum Physics PHY3702 Assignment
T 4 = 1.0141 × 1016 K4
Take the fourth root of both sides to isolate T :
1/4
T = 1.0141 × 1016
Break this into manageable parts:
1/4
T = (1.0141)1/4 × 1016
T = 1.0035 × 104
T ≈ 1.00 × 104 K
1.2(b) Wavelength of Strongest Radiation
Wien’s displacement law connects the peak emission wavelength λmax to the absolute tempera-
ture:
λmax T = b
where b = 2.898 × 10−3 m K is Wien’s displacement constant.
Rearrange to solve for λmax :
b
λmax =
T
Substitute T = 1.00 × 104 K:
2.898 × 10−3
λmax =
1.00 × 104
Page 2 of 35
,UNISA | Quantum Physics PHY3702 Assignment
λmax = 2.898 × 10−7 m
λmax ≈ 2.90 × 10−7 m = 290 nm
Implementation Insight
This wavelength falls in the near-ultraviolet region of the electromagnetic spectrum.
Stars with surface temperatures around 10,000 K (such as A-type stars including Sirius
and Vega) peak in the ultraviolet, which explains why they appear blue-white in visible
light.
Page 3 of 35
,UNISA | Quantum Physics PHY3702 Assignment
Question 2: Photoelectric Effect — Work Function, Cutoff Wavelength and Max-
imum Kinetic Energy
Question: If the stopping potential of a metal when illuminated with radiation of wavelength
480 nm is 1.2 V, find:
1. the work function of the metal,
2. the cutoff wavelength of the metal, and
3. the maximum energy of the ejected electrons.
2.1 Given Information and Photon Energy
λ = 480 nm = 480 × 10−9 m, Vs = 1.2 V
Constants used:
h = 6.626 × 10−34 J s, c = 3.0 × 108 m s−1
Calculate the energy of the incident photon using E = hc/λ:
(6.626 × 10−34 )(3.0 × 108 )
E=
480 × 10−9
1.9878 × 10−25
E=
4.80 × 10−7
E = 4.141 × 10−19 J
Convert to electron-volts by dividing by e = 1.6 × 10−19 C:
4.141 × 10−19
E= = 2.59 eV
1.6 × 10−19
2.2(a) Work Function
The photoelectric equation states:
Page 4 of 35
, UNISA | Quantum Physics PHY3702 Assignment
E = ϕ + Kmax
The stopping potential Vs directly gives the maximum kinetic energy of ejected electrons:
Kmax = eVs = (1)(1.2) eV = 1.2 eV
Therefore, rearrange to find ϕ:
ϕ = E − Kmax
ϕ = 2.59 − 1.2
ϕ = 1.39 eV
2.2(b) Cutoff Wavelength
The cutoff wavelength λ0 is the longest wavelength that can eject electrons. At this wave-
length, all the photon energy goes into overcoming the work function (Kmax = 0):
hc
λ0 =
ϕ
First convert ϕ from eV to joules:
ϕ = 1.39 × (1.6 × 10−19 ) = 2.224 × 10−19 J
Then:
(6.626 × 10−34 )(3.0 × 108 )
λ0 =
2.224 × 10−19
Page 5 of 35
College of Science, Engineering and Technology
⋄
PHY3702 ASSIGNMENT
Modern Physics — Blackbody Radiation, Photoelectric Effect, Compton Scattering,
de Broglie Wavelength, Linear Algebra and Quantum Mechanics
⋄
Module: Quantum Physics
Moodule Code: PHY3702
Assignment No.: 1
Due Date: May 2026
Submitted in partial fulfilment of the requirements for Modern Physics
at the University of South Africa.
,UNISA | Quantum Physics PHY3702 Assignment
Question 1: Blackbody Radiation — Stefan-Boltzmann Law and Wien’s Displace-
ment Law
Question: Assuming that a given star radiates like a blackbody, estimate:
1. the temperature at its surface,
2. the wavelength of its strongest radiation,
when it emits a total intensity of I = 575 MW m−2 .
1.1 Given Information
I = 575 MW m−2 = 575 × 106 W m−2
The Stefan-Boltzmann law governs the total power radiated per unit area by a blackbody:
I = σT 4
where the Stefan-Boltzmann constant is:
σ = 5.67 × 10−8 W m−2 K−4
1.2(a) Surface Temperature
Rearrange the Stefan-Boltzmann law to solve for T 4 :
I
T4 =
σ
Substitute the known values:
575 × 106
T4 =
5.67 × 10−8
5.75 × 108
T4 =
5.67 × 10−8
Page 1 of 35
,UNISA | Quantum Physics PHY3702 Assignment
T 4 = 1.0141 × 1016 K4
Take the fourth root of both sides to isolate T :
1/4
T = 1.0141 × 1016
Break this into manageable parts:
1/4
T = (1.0141)1/4 × 1016
T = 1.0035 × 104
T ≈ 1.00 × 104 K
1.2(b) Wavelength of Strongest Radiation
Wien’s displacement law connects the peak emission wavelength λmax to the absolute tempera-
ture:
λmax T = b
where b = 2.898 × 10−3 m K is Wien’s displacement constant.
Rearrange to solve for λmax :
b
λmax =
T
Substitute T = 1.00 × 104 K:
2.898 × 10−3
λmax =
1.00 × 104
Page 2 of 35
,UNISA | Quantum Physics PHY3702 Assignment
λmax = 2.898 × 10−7 m
λmax ≈ 2.90 × 10−7 m = 290 nm
Implementation Insight
This wavelength falls in the near-ultraviolet region of the electromagnetic spectrum.
Stars with surface temperatures around 10,000 K (such as A-type stars including Sirius
and Vega) peak in the ultraviolet, which explains why they appear blue-white in visible
light.
Page 3 of 35
,UNISA | Quantum Physics PHY3702 Assignment
Question 2: Photoelectric Effect — Work Function, Cutoff Wavelength and Max-
imum Kinetic Energy
Question: If the stopping potential of a metal when illuminated with radiation of wavelength
480 nm is 1.2 V, find:
1. the work function of the metal,
2. the cutoff wavelength of the metal, and
3. the maximum energy of the ejected electrons.
2.1 Given Information and Photon Energy
λ = 480 nm = 480 × 10−9 m, Vs = 1.2 V
Constants used:
h = 6.626 × 10−34 J s, c = 3.0 × 108 m s−1
Calculate the energy of the incident photon using E = hc/λ:
(6.626 × 10−34 )(3.0 × 108 )
E=
480 × 10−9
1.9878 × 10−25
E=
4.80 × 10−7
E = 4.141 × 10−19 J
Convert to electron-volts by dividing by e = 1.6 × 10−19 C:
4.141 × 10−19
E= = 2.59 eV
1.6 × 10−19
2.2(a) Work Function
The photoelectric equation states:
Page 4 of 35
, UNISA | Quantum Physics PHY3702 Assignment
E = ϕ + Kmax
The stopping potential Vs directly gives the maximum kinetic energy of ejected electrons:
Kmax = eVs = (1)(1.2) eV = 1.2 eV
Therefore, rearrange to find ϕ:
ϕ = E − Kmax
ϕ = 2.59 − 1.2
ϕ = 1.39 eV
2.2(b) Cutoff Wavelength
The cutoff wavelength λ0 is the longest wavelength that can eject electrons. At this wave-
length, all the photon energy goes into overcoming the work function (Kmax = 0):
hc
λ0 =
ϕ
First convert ϕ from eV to joules:
ϕ = 1.39 × (1.6 × 10−19 ) = 2.224 × 10−19 J
Then:
(6.626 × 10−34 )(3.0 × 108 )
λ0 =
2.224 × 10−19
Page 5 of 35
