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Physics 1201 Final Exam Review: Electricity, Magnetism, Optics, and Quantum Mechanics, Exams of Physics

A comprehensive review of physics 1201, covering lectures 1 through 27. It includes key topics such as electric force, electric fields, electric potential, electric currents, dc circuits, magnetic forces and fields, electromagnetic induction, electromagnetic waves, reflection and refraction of light, interference, polarization, blackbody radiation, the photoelectric effect, the bohr atom, heisenberg uncertainty principle, and nuclear physics. The review also features examples and explanations of malus law, wiens displacement law, and the de broglie wavelength, making it a valuable resource for students preparing for their final exam. It also covers the mass deficit of the nucleus and nuclear binding energy. Useful for university students.

Typology: Exams

2024/2025

Available from 05/30/2025

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Physics 1201 Final exam review
Lectures 1 27
ØMonday, 4/28/25 from 6:00 pm 7:45 pm
ØComprehensive covers all material in course
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Download Physics 1201 Final Exam Review: Electricity, Magnetism, Optics, and Quantum Mechanics and more Exams Physics in PDF only on Docsity!

Physics 1201 Final exam review

Lectures 1 – 27

Ø Monday, 4/28/25 from 6:00 pm – 7:45 pm

Ø Comprehensive – covers all material in course

Physics 1201 Midterm 1 review

Electric force

Electric field

Electric potential

Electric currents

DC circuits and circuit devices

Lectures 1 – 9 Electricity

à See Midterm 1 review (Lecture 12)

Physics 1201 review since MT 2

Lectures 20 – 27

Polarization of waves

Blackbody radiation

Photoelectric effect

Bohr atom

Heisenberg uncertainty principle

Nuclear physics

Radiation effects

Chapter 24

Interference and the

Wave Nature of Light

Lecture 20

Polarization Example: Using Polarizers and Analyzers What value of q should be used so the average intensity of the po light reaching the photocell is one-tenth the average intensity of th unpolarized light?

Polarization

= cos

2 θ cos^ θ^ =^

 θ = 63. 4 Applying Malus’ Law at the Analyzer: But and we want So I 0 I 0,polar I è è 𝐼 = 𝐼!,#$%&' cos 2 𝜃 I0,polar = ( ) I 0 𝐼 = 1 10 𝐼! 1 10 𝐼! = 1 2 𝐼! cos 2 𝜃

q Blackbody radiation

  • The nature of the blackbody radiation depends only on the temperature of the body, not on the material composition of the object.
  • The distribution of energy in blackbody radiation varies with wavelength and temperature. - The total amount of energy (area under the curve) it emits increase with the temperature. - The peak of the distribution shifts to shorter wavelengths. This shift obeys Wiens displace- ment law:

max

T = 2. 898 × 10

− 3 m ⋅ K

  • The classical theory of thermal radiation at the end of 19 th century failed to explain the distribution of energy of the blackbody radiation. Blackbody radiation

Example. a) Find the temperature of the Sun’s surface assuming that it emits blackbody radiation and the wavelength of the peak in its intensity distribution is at 500 nm (in the yellow region).b) Find the temperature of the star Betelgeuse (in the Orion constellation) which is classified as a Red Giant (near the end of its life cycle) with peak wavelength 890 nm. λ max

T = 2. 898 × 10

− 3 m ⋅ K T =

2. 898 × 10

− 3 λ max

2. 898 × 10

− 3 500 × 10 − 9

= 5800 K
T =
2. 898 × 10

− 3 890 × 10 − 9 = 3260 K Much cooler than the Sun a) Sun: b) Betelgeuse: Blackbody radiation

Example. a) Find the energy of a photon with wavelength 210 nm. b) When light with this wavelength falls on a metal, the photoelectric circuit is brought to zero at a stopping voltage of 1.64 V. Find the work function of the metal. a ) E = hf = h c λ

= 6. 63 × 10

− 34 ( )

3. 00 × 10

8 ( ) 210 × 10 − 9 ( )

= 9. 47 × 10

− 19 J = 5. 92 eV b ) KE max = e Δ V s

= 1. 60 × 10

− 19 ( ) (^1.^64 )^ =^2.^62 ×^10 − 19 J = 1. 64 eV KE max = hfW 0

⇒ W

0 = hfKE max = 5. 92 − 1. 64 = 4. 28 eV The Photoelectric Effect

Example. What are the a) momentum and b) energy of a blue photon with wavelength 470 nm?

a ) p =

h

λ

6. 63 × 10

− 34

470 × 10

− 9

= 1. 41 × 10

− 27

kg

m

s

b ) E = pc = 1. 41 × 10

− 27 ( )

3. 00 × 10

8 ( )

= 4. 23 × 10

− 19

J = 2. 64 eV

The Photoelectric Effect

Example. What kinetic energy should neutrons (m = 1.67 x 10

  • 27 kg) have to exhibit diffraction effects in a crystal with interatomic spacing of 1.00 x 10
  • 10 m? λ = h pp = h λ
6. 63 × 10

− 34

  1. 00 × 10 − 10
= 6. 63 × 10

− 24 kg m s KE = p 2 2 m

6. 63 × 10

− 24

2 2 1. 67 × 10 − 27

= 1. 32 × 10

− 20 J = 0. 0825 eV e.g. “thermal” neutrons from a nuclear reactor corresponding to a temperature of 640 o K KE = 3 2 kT k = Boltzman const. Set the neutron wavelength to be the interatomic spacing: l = 1.00 x 10

  • 10 m de Broglie wavelength

The Bohr Model of the Hydrogen Atom 1 , 2 , 3 ,  2 = = n = h L mv r n n n n p ∴ r n = h 2 4 π 2 mke 2 "

$ % & ' n 2 Z n = 1 , 2 , 3 , … ( 5. 29 10 m) 1 , 2 , 3 , 2 11 = ´ =

  • n Z n r n Radii for Bohr orbits Angular momentum is quantized. We can get rn by combining this with (from the previous slide), mv 2 r = F centrip = kZe 2 r 2 or , m 2 v n 2 = mkZe 2 r n Bohr radius

Example: Find the wavelength of the second Balmer line in hydrogen, i.e. the n i = 4 to n f = 2 transition. 1 λ = ( 1. 097 × 10 7 m − 1 ) 1 n f 2 − 1 n i 2

$ % % & ' ( ( = ( 1. 097 × 10 7 m − 1 ) 1 2 2 − 1 4 2

$ % & ' ( =^2.^057 ×^10 6 m − 1 ⇒ λ = 486 nm blue The Bohr Model of the Hydrogen Atom

Example: Ionization of a hydrogen atom. What wavelength photon is needed to ionize a hydrogen atom in the ground state and give the ejected electron 11.0 eV of kinetic energy? KE = hfE 1 hc λ

= KE + E

1 = 11. 0 + 13. 6 = 24. 6 eV = 3. 94 × 10 − 18 J λ = hc

  1. 94 × 10 − 18
6. 63 × 10

− 34 ( ) 3.^00 ×^10 8 ( )

  1. 94 × 10 − 18 = 5. 05 × 10 − 8 m = 50. 5 nm ultraviolet The Bohr Model of the Hydrogen Atom ( 13. 6 eV) (^1) , 2 , 3 , 2 2 = - n = n Z En