AP Physics Practice Questions: Modern Physics

AP Physics multiple choice questions on quantum mechanics, relativity, and nuclear physics. Each question includes detailed solutions at the bottom of the page.

1. What is the de Broglie wavelength of an electron with kinetic energy 100 eV? (h = 6.63×10⁻³⁴ J·s, m_e = 9.11×10⁻³¹ kg)

   A) 0.123 nm
   B) 0.387 nm
   C) 1.23 nm
   D) 3.87 nm
   E) 12.3 nm

2. A spaceship travels at 0.8c relative to Earth. If the spaceship measures a time interval of 1 hour, what time interval is measured on Earth?

   A) 0.6 hours
   B) 0.8 hours
   C) 1.0 hour
   D) 1.25 hours
   E) 1.67 hours

3. Light of wavelength 400 nm is incident on a metal with work function 2.0 eV. What is the maximum kinetic energy of emitted electrons?

   A) 0.1 eV
   B) 1.1 eV
   C) 3.1 eV
   D) 5.1 eV
   E) 7.1 eV

4. What is the rest energy of a proton? (m_p = 1.67×10⁻²⁷ kg, c = 3×10⁸ m/s)

   A) 1.50×10⁻¹⁰ J
   B) 1.67×10⁻¹⁰ J
   C) 9.38×10⁻¹⁰ J
   D) 1.50×10⁻⁹ J
   E) 1.67×10⁻⁹ J

5. ²³⁸U undergoes alpha decay. What is the resulting nucleus?

   A) ²³⁴Th
   B) ²³⁴U
   C) ²³⁴Pa
   D) ²³⁶Th
   E) ²³⁶U

Solutions

1. A) 0.123 nm
   First convert 100 eV to joules: \(100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17}\,\text{J}\)
   Kinetic energy: \(K = \frac{p^2}{2m} \Rightarrow p = \sqrt{2 m K}\)
   \(p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-17}} = \sqrt{2.915 \times 10^{-47}} = 5.40 \times 10^{-24}\,\text{kg·m/s}\)
   De Broglie wavelength: \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.228 \times 10^{-10}\,\text{m} = 0.123\,\text{nm}\)

2. E) 1.67 hours
   Time dilation: \(\Delta t = \frac{\Delta t_0}{\sqrt{1 - v^2/c^2}}\)
   \(\Delta t = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = 1.67\,\text{hours}\)

3. B) 1.1 eV
   Energy of photon: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}}\)
   \(E = \frac{1.989 \times 10^{-25}}{400 \times 10^{-9}} = 4.97 \times 10^{-19}\,\text{J}\)
   Convert to eV: \(\frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.11\,\text{eV}\)
   Maximum kinetic energy: \(K_{\text{max}} = E - \phi = 3.11 - 2.0 = 1.11\,\text{eV}\)

4. A) 1.50×10⁻¹⁰ J
   \(E = mc^2 = 1.67 \times 10^{-27} \times (3 \times 10^8)^2 = 1.67 \times 10^{-27} \times 9 \times 10^{16} = 1.503 \times 10^{-10}\,\text{J}\)

5. A) \(^ {234}\text{Th}\)
   Alpha decay: \(^ {238}\text{U} \rightarrow ^4\text{He} + ^{234}\text{Th}\)
   Mass number: \(238 - 4 = 234\)
   Atomic number: \(92 - 2 = 90\) (Thorium)