Quantum Physics
PHY3702
Year Module
Department of Physics
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,Question 1
Given
𝐼 = 575 MW/m2 = 5.75 × 108 W/m2
𝜎 = 5.67 × 10−8 W/m2 K4
𝑏 = 2.898 × 10−3 m.K
(a) Surface temperature
Using Stefan–Boltzmann law:
𝐼 = 𝜎𝑇 4
𝐼 1/4
𝑇=( )
𝜎
Substitute:
1/4
5.75 × 108
𝑇=( )
5.67 × 10−8
𝑇 = (1.014 × 1016 )1/4
𝑇 ≈ 1.00 × 104 K
(b) Wavelength of strongest radiation
Using Wien’s law:
𝑏
𝜆max =
𝑇
2.898 × 10−3
𝜆max =
1.00 × 104
𝜆max ≈ 2.90 × 10−7 m
Final Answers
𝑇 ≈ 1.0 × 104 K
𝜆max ≈ 2.9 × 10−7 m (290 nm)
, Question 2
Given
𝜆 = 480 nm = 4.80 × 10−7 m
𝑉𝑠 = 1.2 V
ℎ = 6.63 × 10−34 J.s
𝑐 = 3.00 × 108 m/s
𝑒 = 1.60 × 10−19 C
(a) Work function
Photon energy
ℎ𝑐
𝐸=
𝜆
(6.63 × 10−34 )(3.00 × 108 )
𝐸=
4.80 × 10−7
𝐸 ≈ 4.14 × 10−19 J
Kinetic energy
𝐾max = 𝑒𝑉𝑠 = (1.60 × 10−19 )(1.2)
𝐾max = 1.92 × 10−19 J
Work function
𝜙 = 𝐸 − 𝐾max
𝜙 = 4.14 × 10−19 − 1.92 × 10−19
𝜙 = 2.22 × 10−19 J
(b) Cutoff wavelength
ℎ𝑐
𝜆0 =
𝜙
(6.63 × 10−34 )(3.00 × 108 )
𝜆0 =
2.22 × 10−19
𝜆0 ≈ 8.96 × 10−7 m