Dawn of modern physics

NUST NET (Engineering) · Physics · Modern and nuclear physics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Three ideas changed physics in the early 1900s. First, special relativity: the speed of light is the same for every observer, so a moving clock runs slow, a moving object is shorter along its motion and its mass rises with speed. The effects matter only near the speed of light.

Second, light arrives in packets called photons, each of energy hf. One photon gives its energy to one electron, which explains the photoelectric effect: frequency decides the energy of the electrons, intensity decides only how many are emitted. The Compton effect shows that photons carry momentum too.

Third, particles behave as waves with a de Broglie wavelength, and position and momentum cannot both be known exactly.

Rules to remember

Common mistake

Thinking brighter light gives faster photoelectrons. Intensity changes only the number emitted each second; the maximum kinetic energy depends on the frequency alone.

Worked example

Light of wavelength 310 nm falls on a metal of work function 2.5 eV. Taking hc = 1240 eV nm, what is the maximum kinetic energy of the photoelectrons?

  1. Photon energy = hc/λ = 1240/310 = 4.0 eV.
  2. This is more than the work function, so electrons are emitted.
  3. KEmax = hf − Φ = 4.0 − 2.5 = 1.5 eV.

Answer: 1.5 eV (so the stopping potential is 1.5 V).

Practice questions

Try each one, then open the answer.

1. An electron is accelerated from rest through a potential difference V. If V is made four times as large, its de Broglie wavelength

  1. A
    doubles
  2. B
    halves
  3. C
    becomes one-quarter
  4. D
    is unchanged
Show answer

Answer: B. eV = p2/2m, so p ∝ √V and λ = h/p ∝ 1/√V. Four times V doubles p and halves λ. One-quarter wrongly takes λ ∝ 1/V.

2. The minimum energy of a photon that can create an electron–positron pair is about

  1. A
    0.51 MeV
  2. B
    2.04 MeV
  3. C
    931 MeV
  4. D
    1.02 MeV
Show answer

Answer: D. The photon must supply the rest energy of two particles: 2m0c2 = 2 × 0.51 MeV = 1.02 MeV. 0.51 MeV is the rest energy of only one electron.

3. Light of photon energy 5.0 eV falls on a metal of work function 2.0 eV. The stopping potential for the emitted photoelectrons is

  1. A
    2.0 V
  2. B
    2.5 V
  3. C
    3.0 V
  4. D
    7.0 V
Show answer

Answer: C. Kmax = hf − φ = 5.0 − 2.0 = 3.0 eV, and eV0 = Kmax, so V0 = 3.0 V. 7.0 V adds the work function instead of subtracting it.

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