Quantum Mechanics — Practice Quiz

A Physics cheat sheet for Quantum Mechanics — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.

Formulas & key concepts

Planck's Quantized Energies: Energy \(E\) of an atomic oscillator is quantized, where \(n\) is an integer, \(h\) is Planck's constant, and \(f\) is frequency.

$$E = nhf$$

Where: \(n = 0, 1, 2, \dots\); \(h = 6.63 \times 10^{-34} J\cdot s\)

Photon Energy: The energy \(E\) of a single photon is proportional to its frequency \(f\).

$$E = hf$$

Where: \(h\) = Planck's constant

Photoelectric Effect: Energy conservation where photon energy \(hf\) equals maximum kinetic energy \(KE_{max}\) of ejected electron plus work function \(W_0\).

$$hf = KE_{max} + W_0$$

Photon Momentum: Momentum \(p\) of a photon related to its wavelength \(\lambda\).

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

Compton Effect: Shift in wavelength \(\lambda' - \lambda\) when a photon scatters off an electron of mass \(m\) at angle \(\theta\).

$$\lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta)$$

Where: \(h/mc\) = Compton wavelength of electron

De Broglie Wavelength: Wavelength \(\lambda\) associated with a particle of momentum \(p\).

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

Heisenberg Uncertainty Principle (Position-Momentum): Fundamental limit on precision of position \(\Delta y\) and momentum \(\Delta p_y\).

$$(\Delta p_y)(\Delta y) \ge \frac{h}{4\pi}$$

Heisenberg Uncertainty Principle (Energy-Time): Fundamental limit on precision of energy \(\Delta E\) and time interval \(\Delta t\).

$$(\Delta E)(\Delta t) \ge \frac{h}{4\pi}$$

Practice quiz

  1. An atomic oscillator has a frequency of $5.0 \times 10^{14} \text{ Hz}$. What is the energy of its first excited state ($n=1$)? Use Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$.

    • $3.315 \times 10^{-19} \text{ J}$
    • $6.63 \times 10^{-34} \text{ J}$
    • $1.658 \times 10^{-19} \text{ J}$
    • $0 \text{ J}$

    Answer: $3.315 \times 10^{-19} \text{ J}$

  2. A photon has a wavelength of $500 \text{ nm}$. What is its energy in electron volts ($ \text{eV}$)? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$, $c = 3.00 \times 10^8 \text{ m/s}$, and $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$.

    • $2.49 \text{ eV}$
    • $1.24 \text{ eV}$
    • $3.98 \times 10^{-19} \text{ J}$
    • $6.00 \times 10^{14} \text{ Hz}$

    Answer: $2.49 \text{ eV}$

  3. Light with a frequency of $1.0 \times 10^{15} \text{ Hz}$ shines on a metal surface. If the work function of the metal is $2.5 \text{ eV}$, what is the maximum kinetic energy of the ejected electrons? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$ and $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$.

    • $1.64 \text{ eV}$
    • $2.5 \text{ eV}$
    • $4.13 \text{ eV}$
    • $0 \text{ eV}$

    Answer: $1.64 \text{ eV}$

  4. What is the momentum of a photon with a wavelength of $663 \text{ nm}$? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$.

    • $1.00 \times 10^{-27} \text{ kg} \cdot \text{m/s}$
    • $1.00 \times 10^{-25} \text{ kg} \cdot \text{m/s}$
    • $6.63 \times 10^{-34} \text{ kg} \cdot \text{m/s}$
    • $3.00 \times 10^8 \text{ kg} \cdot \text{m/s}$

    Answer: $1.00 \times 10^{-27} \text{ kg} \cdot \text{m/s}$

  5. In the Compton effect, if a photon scatters off an electron at an angle of $90$ degrees, what is the change in its wavelength?

    • The Compton wavelength of the electron.
    • Zero, as the photon loses no energy.
    • Twice the Compton wavelength of the electron.
    • Dependent on the initial wavelength of the photon.

    Answer: The Compton wavelength of the electron.

  6. An electron is accelerated through a potential difference of $100 \text{ V}$. What is its de Broglie wavelength? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$, $m_e = 9.11 \times 10^{-31} \text{ kg}$, and $e = 1.60 \times 10^{-19} \text{ C}$.

    • $1.23 \times 10^{-10} \text{ m}$
    • $6.63 \times 10^{-34} \text{ m}$
    • $1.00 \times 10^{-10} \text{ m}$
    • $5.92 \times 10^6 \text{ m}$

    Answer: $1.23 \times 10^{-10} \text{ m}$

  7. If the position of an electron can be determined with an uncertainty of $1.0 \text{ nm}$, what is the minimum uncertainty in its momentum? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$.

    • $5.28 \times 10^{-26} \text{ kg} \cdot \text{m/s}$
    • $1.05 \times 10^{-25} \text{ kg} \cdot \text{m/s}$
    • $6.63 \times 10^{-34} \text{ kg} \cdot \text{m/s}$
    • $0 \text{ kg} \cdot \text{m/s}$

    Answer: $5.28 \times 10^{-26} \text{ kg} \cdot \text{m/s}$

  8. An excited state of an atom has a lifetime of $1.0 \times 10^{-8} \text{ s}$. What is the minimum uncertainty in the energy of this state? Use $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$.

    • $5.28 \times 10^{-27} \text{ J}$
    • $1.05 \times 10^{-26} \text{ J}$
    • $6.63 \times 10^{-34} \text{ J}$
    • $0 \text{ J}$

    Answer: $5.28 \times 10^{-27} \text{ J}$

  9. A photon has the same momentum as an electron moving at $1.0 \times 10^6 \text{ m/s}$. What is the energy of this photon? Use $m_e = 9.11 \times 10^{-31} \text{ kg}$ and $c = 3.00 \times 10^8 \text{ m/s}$.

    • $2.73 \times 10^{-16} \text{ J}$
    • $9.11 \times 10^{-25} \text{ J}$
    • $6.63 \times 10^{-34} \text{ J}$
    • $3.00 \times 10^8 \text{ J}$

    Answer: $2.73 \times 10^{-16} \text{ J}$

  10. For the photoelectric effect to occur, the incident photon energy $hf$ must be:

    • Greater than or equal to the work function $W_0$.
    • Less than the work function $W_0$.
    • Equal to the maximum kinetic energy $KE_{max}$.
    • Independent of the work function $W_0$.

    Answer: Greater than or equal to the work function $W_0$.

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