FAD1022 Tutorial 15 — Modern Physics (Student Version)

Course: FAD1022 Basic Physics II
Semester: 2025/2026
Centre: Centre for Foundation Studies in Science
Lecturer: Nurul Izzati (NIA)

Constants Provided

Constant Symbol Value
Planck's constant $h$ $6.63 \times 10^{-34}$ J·s
Elementary charge $e$ $1.60 \times 10^{-19}$ C
Speed of light $c$ $3.0 \times 10^{8}$ m/s
Electron mass $m_e$ $9.11 \times 10^{-31}$ kg

Question 1: Wave-Particle Duality

Question: State the wave-particle duality and explain why we do not observe the wave nature of large moving objects like a human.

Answer

Wave-particle duality states that all matter and light exhibit both wave-like and particle-like properties.

Why we don't observe wave nature in large objects: The de Broglie wavelength is given by: $$\lambda = \frac{h}{mv}$$

For large objects (like humans), the mass $m$ is so large that $\lambda$ becomes astronomically small — far too small to be detected or observed. Only particles with very small mass (like electrons) have measurable wavelengths.

Question 2: Photoelectric Effect (EXAM FAVORITE)

[!warning] This exact problem type appears in almost every exam!

Question: In a photoelectric effect experiment, a metal surface is illuminated with two different light sources as shown in Table 1. Calculate the work function, $\phi$ and Planck's constant, $h$ for this experiment.

Table 1

Light sources 1 2
Frequency $7.0 \times 10^{14}$ Hz $1.0 \times 10^{15}$ Hz
Stopping potential 0.5 V 1.8 V

Solution

Step 1: Write the photoelectric equation for both data points

$$eV_s = hf - \phi$$

For source 1: $$e(0.5) = h(7.0 \times 10^{14}) - \phi$$ $$0.5e = 7.0 \times 10^{14}h - \phi \quad \text{... (1)}$$

For source 2: $$e(1.8) = h(1.0 \times 10^{15}) - \phi$$ $$1.8e = 1.0 \times 10^{15}h - \phi \quad \text{... (2)}$$

Step 2: Subtract equation (1) from equation (2)

$$(1.8 - 0.5)e = (1.0 \times 10^{15} - 7.0 \times 10^{14})h$$

$$1.3e = 3.0 \times 10^{14}h$$

$$h = \frac{1.3 \times 1.6 \times 10^{-19}}{3.0 \times 10^{14}}$$

$$\boxed{h = 6.93 \times 10^{-34} \text{ J·s}}$$

Step 3: Find $\phi$ using equation (1)

$$\phi = 7.0 \times 10^{14}h - 0.5e$$

$$\phi = 7.0 \times 10^{14}(6.93 \times 10^{-34}) - 0.5(1.6 \times 10^{-19})$$

$$\phi = 4.85 \times 10^{-19} - 0.8 \times 10^{-19}$$

$$\boxed{\phi = 4.05 \times 10^{-19} \text{ J} = 2.53 \text{ eV}}$$

Question 3: Photon Energy and Momentum

Question: In a medical phototherapy treatment, a specialized laser emits photons that each carry an energy of $4.5 \times 10^{-19}$ J. Calculate the wavelength of this medical laser and determine the momentum transferred by a single photon to the targeted skin tissue.

Solution

Part (a): Wavelength

$$E = \frac{hc}{\lambda}$$

$$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{4.5 \times 10^{-19}}$$

$$\boxed{\lambda = 4.42 \times 10^{-7} \text{ m} = 442 \text{ nm}}$$

Part (b): Photon Momentum

$$p = \frac{h}{\lambda} = \frac{E}{c}$$

$$p = \frac{4.5 \times 10^{-19}}{3.0 \times 10^{8}}$$

$$\boxed{p = 1.5 \times 10^{-27} \text{ kg·m/s}}$$

Question 4: de Broglie Wavelength

Question: Calculate the de-Broglie wavelength of an electron that has a kinetic energy of 200 eV.

Solution

Step 1: Convert KE to joules

$$KE = 200 \text{ eV} = 200 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-17} \text{ J}$$

Step 2: Use the de Broglie formula

$$\lambda = \frac{h}{\sqrt{2m_e KE}}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2(9.11 \times 10^{-31})(3.2 \times 10^{-17})}}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{5.83 \times 10^{-47}}}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{7.64 \times 10^{-24}}$$

$$\boxed{\lambda = 8.68 \times 10^{-11} \text{ m}}$$

Question 5: Heisenberg Uncertainty Principle

Question: If the uncertainty in position of an electron is $1.0 \times 10^{-10}$ m, calculate the minimum uncertainty in its momentum.

Solution

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Using the minimum equality:

$$\Delta p = \frac{h}{4\pi \cdot \Delta x}$$

$$\Delta p = \frac{6.63 \times 10^{-34}}{4\pi(1.0 \times 10^{-10})}$$

$$\boxed{\Delta p = 5.28 \times 10^{-25} \text{ kg·m/s}}$$

Key Formulas Summary

Topic Formula
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mKE}}$
Photon energy $E = hf = \frac{hc}{\lambda}$
Photon momentum $p = \frac{h}{\lambda} = \frac{E}{c}$
Photoelectric effect $eV_s = hf - \phi$
Heisenberg uncertainty $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$

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