Quantum Mechanics — Hard Practice Quiz
A Physics cheat sheet for Quantum Mechanics — every key formula with its symbols defined — plus a hard-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.
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 \(f\).
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 \(W_0\).
Photon Momentum: Momentum \(p\) of a photon related to its wavelength \(\lambda\).
Compton Effect: Shift in wavelength \(\lambda' - \lambda\) when a photon scatters off an electron of mass \(m\) at angle \(\theta\).
Where: \(h/mc\) = Compton wavelength of electron
De Broglie Wavelength: Wavelength \(\lambda\) associated with a particle of momentum \(p\).
Heisenberg Uncertainty Principle (Position-Momentum): Fundamental limit on precision of position \(\Delta y\) and momentum \(\Delta p_y\).
Heisenberg Uncertainty Principle (Energy-Time): Fundamental limit on precision of energy \(\Delta E\) and time interval \(\Delta t\).
Practice quiz
A photon has an energy $E$. If a particle has the same momentum as this photon, what is the de Broglie wavelength of the particle in terms of $E$, Planck's constant $h$, and the speed of light $c$?
- $hc/E$
- $hE/c$
- $h/E$
- $c/E$
Answer: $hc/E$
Light of wavelength $\lambda$ is incident on a metal surface with work function $W_0$. If an electron is ejected, what is its de Broglie wavelength? Assume the electron's speed is non-relativistic and its mass is $m$.
- $h / \sqrt{2m (hc/\lambda - W_0)}$
- $h / \sqrt{2m (hc/\lambda + W_0)}$
- $h \sqrt{2m (hc/\lambda - W_0)}$
- $h / \sqrt{m (hc/\lambda - W_0)}$
Answer: $h / \sqrt{2m (hc/\lambda - W_0)}$
A photon with initial energy $E$ undergoes Compton scattering off a stationary electron at an angle $\theta$. What is the energy of the scattered photon, $E'$, in terms of $E$, $\theta$, Planck's constant $h$, electron mass $m$, and speed of light $c$?
- $\frac{E mc^2}{mc^2 + E(1 - \cos \theta)}$
- $\frac{E mc^2}{mc^2 - E(1 - \cos \theta)}$
- $\frac{E}{1 + \frac{E}{mc^2}(1 - \cos \theta)}$
- $\frac{E mc^2}{E + mc^2(1 - \cos \theta)}$
Answer: $\frac{E mc^2}{mc^2 + E(1 - \cos \theta)}$
If the uncertainty in a particle's position, $\Delta y$, is exactly equal to its de Broglie wavelength $\lambda$, what is the minimum uncertainty in its momentum, $\Delta p_y$, expressed as a fraction of its actual momentum $p$?
- $1/(4\pi)$
- $1/(2\pi)$
- $1/\pi$
- $1$
Answer: $1/(4\pi)$
An atomic oscillator transitions from an energy level $n_1$ to $n_2$, where $n_1 > n_2$. If the frequency of the emitted photon is $f_{photon}$, what is the fundamental frequency $f$ of the oscillator in terms of $f_{photon}$, $n_1$, and $n_2$?
- $\frac{f_{photon}}{n_1 - n_2}$
- $f_{photon}(n_1 - n_2)$
- $\frac{f_{photon}}{n_1 + n_2}$
- $hf_{photon}(n_1 - n_2)$
Answer: $\frac{f_{photon}}{n_1 - n_2}$
A photon has momentum $p_{photon}$. An electron has the same kinetic energy as the photon's energy. What is the de Broglie wavelength of the electron in terms of $p_{photon}$, Planck's constant $h$, electron mass $m_e$, and speed of light $c$?
- $h / \sqrt{2m_e p_{photon} c}$
- $h \sqrt{2m_e p_{photon} c}$
- $h / \sqrt{m_e p_{photon} c}$
- $h / (2m_e p_{photon} c)$
Answer: $h / \sqrt{2m_e p_{photon} c}$
An excited state of an atom has a mean lifetime $\Delta t$. If it decays by emitting a photon, what is the minimum uncertainty in the photon's wavelength, $\Delta \lambda$, in terms of $\Delta t$, Planck's constant $h$, the speed of light $c$, and the photon's central wavelength $\lambda$?
- $\frac{\lambda^2}{4\pi c \Delta t}$
- $\frac{h \lambda^2}{4\pi c \Delta t}$
- $\frac{4\pi c \Delta t}{\lambda^2}$
- $\frac{\lambda}{4\pi c \Delta t}$
Answer: $\frac{\lambda^2}{4\pi c \Delta t}$
A metal surface has a work function $W_0$. When light of frequency $f$ is incident, the maximum kinetic energy of the ejected electrons is $KE_{max}$. If the frequency of the incident light is doubled to $2f$, what is the new maximum kinetic energy, $KE'_{max}$, in terms of $KE_{max}$, $W_0$, and $f$?
- $2KE_{max} + W_0$
- $2KE_{max} - W_0$
- $2KE_{max} + 2W_0$
- $KE_{max} + W_0$
Answer: $2KE_{max} + W_0$
A photon of initial wavelength $\lambda$ scatters off a stationary electron at an angle $\theta$. What is the kinetic energy of the recoiling electron in terms of $\lambda$, $\theta$, Planck's constant $h$, electron mass $m$, and speed of light $c$?
- $\frac{h^2 (1 - \cos \theta)}{m \lambda (\lambda + \frac{h}{mc}(1 - \cos \theta))}$
- $\frac{h^2 (1 + \cos \theta)}{m \lambda (\lambda + \frac{h}{mc}(1 - \cos \theta))}$
- $\frac{h^2 (1 - \cos \theta)}{m \lambda (\lambda - \frac{h}{mc}(1 - \cos \theta))}$
- $\frac{h (1 - \cos \theta)}{m \lambda (\lambda + \frac{h}{mc}(1 - \cos \theta))}$
Answer: $\frac{h^2 (1 - \cos \theta)}{m \lambda (\lambda + \frac{h}{mc}(1 - \cos \theta))}$
If a particle's momentum $p$ is known with perfect certainty (i.e., $\Delta p_y = 0$), what can be concluded about its de Broglie wavelength $\lambda$ and its position uncertainty $\Delta y$?
- Its de Broglie wavelength is precisely defined, and its position uncertainty is infinite.
- Its de Broglie wavelength is infinite, and its position uncertainty is zero.
- Both its de Broglie wavelength and position uncertainty are precisely defined.
- Both its de Broglie wavelength and position uncertainty are infinite.
Answer: Its de Broglie wavelength is precisely defined, and its position uncertainty is infinite.
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