Foundations of Quantum Mechanics: 18 MCQs with answers

Intensity rises without limit

Classical theory predicted intensity would rise without limit as wavelength decreased — the ultraviolet catastrophe. Experiments showed intensity peaks and then falls off.

$E = h\nu$

Planck's quantum hypothesis states that the energy of one quantum is $E = h\nu$, where $h$ is Planck's constant and $\nu$ is frequency.

No electrons are emitted

Below the threshold frequency, no electrons are emitted regardless of how bright the light is. The photon energy $h\nu$ is insufficient to overcome the work function.

It has wavelength $\lambda = h/p$

de Broglie proposed that every particle has an associated wavelength $\lambda = h/p$, where $h$ is Planck's constant and $p$ is momentum.

Davisson-Germer experiment

The Davisson-Germer experiment (1927) observed electron diffraction from a nickel crystal, providing the first direct proof of matter waves.

True

Planck's constant $h = 6.626 \times 10^{-34}$ J·s and $\hbar = h/2\pi = 1.055 \times 10^{-34}$ J·s are the fundamental constants of quantum mechanics.

Because the uncertainty is a fundamental property of the quantum state

The uncertainty principle is not about measurement error. Even with perfect instruments, a quantum particle genuinely does not have simultaneously sharp position and momentum — the uncertainty is in the state itself.

The work function of the metal

$\phi$ is the work function — the minimum energy an electron needs to escape the metal surface. It is subtracted from the photon energy $h\nu$ to give the maximum kinetic energy.

Photons carry momentum $p = h/\lambda$

Compton treated the interaction as a collision between a photon with momentum $p = h/\lambda$ and a free electron. The wavelength shift matched conservation of energy and momentum exactly.

Always positive

A wave function must be continuous, single-valued, finite, and normalisable. It does not need to be always positive — it is generally complex-valued.

The probability density of finding the particle

$|\Psi|^2$ gives the probability density of finding the particle at a given position. The wave function itself is not directly measurable.

Beam A ejects more electrons, but with the same maximum kinetic energy

Intensity determines the number of photons, hence the number of ejected electrons. The maximum kinetic energy depends only on frequency, not intensity.

Because $\lambda = h/p$ is extremely tiny for large momentum

For everyday objects, momentum $p$ is large, so $\lambda = h/p$ is unimaginably small. For electrons, $\lambda$ is comparable to atomic spacings, making it measurable.

Both A and R are true, and R correctly explains A.

Classical wave theory predicted that brighter light (more energy) should always eject electrons and that kinetic energy should depend on intensity. Experiments showed the opposite — this is exactly why classical theory failed.

I, II, and III

All three statements are correct. The photoelectric effect and Compton scattering demonstrate particle-like behaviour of light, while Davisson-Germer demonstrates wave-like behaviour of electrons.

$\Delta p$ must be very large

By the uncertainty principle $\Delta x \cdot \Delta p \geq \hbar/2$, if $\Delta x$ is very small, $\Delta p$ must be correspondingly large. The product cannot be less than $\hbar/2$.

It doubles.

According to the uncertainty principle, if $$\Delta x$$ is halved, the product $$\Delta x \cdot \Delta p$$ must still satisfy the inequality. Therefore, $$\Delta p$$ must increase to maintain the inequality, specifically doubling in this case.

Zero.

When the photon energy is equal to the work function, the electron has just enough energy to escape the metal surface, resulting in zero kinetic energy after emission, as there is no leftover energy.