Progressive Quantum physics practice questions
Use Clevolab for progressive practice in quantum physics. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats quantum physics as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current Progressive set covers photons, energy levels, uncertainty, and wave-particle ideas. 148 reviewed questions currently published for this page.
What you can practise
- Photon model
- Energy levels
- Wave-particle duality
- Uncertainty ideas
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
At fixed frequency above threshold, increasing light intensity will make the photoelectric current
- AIncreaseCorrect answer
- BDecreaseAnswer option
- CUnchangedAnswer option
- DReverse directionAnswer option
Why this answer is right
Higher intensity means more photons per second. With each photon able to eject an electron, the emission rate (and current) increases. The electron energies do not change at fixed frequency.
Photon flux controls the emission rate when $f>f_0$.
Sample question
For integers $m\ne n$, evaluate $\int_0^L \sin\!\left(\frac{m\pi x}{L}\right)\sin\!\left(\frac{n\pi x}{L}\right) dx$.
- A$0$Correct answer
- B$\frac{L}{n+m}$Answer option
- C$\frac{L}{2}$Answer option
- D$\frac{2L}{\pi}$Answer option
Why this answer is right
Eigenfunctions of the infinite well are orthogonal. The integral of the product of different sine modes over the domain vanishes, so it equals 0 for $m\ne n$.
Orthogonality comes from Sturm–Liouville theory and boundary conditions. For the infinite well on $0<x<L$, $$\int_0^L \sin\!\left(\frac{m\pi x}{L}\right) \sin\!\left(\frac{n\pi x}{L}\right) dx = 0 \quad (m\ne n).$$
Sample question
Which electron arrangement violates the Pauli exclusion principle?
- ATwo electrons in $2p$ with $m_\ell=0$, $m_s=+\tfrac{1}{2}$ and $m_\ell=+1$, $m_s=+\tfrac{1}{2}$Answer option
- BTwo electrons in $2p$ with $m_\ell=0$, $m_s=+\tfrac{1}{2}$ and $m_\ell=0$, $m_s=-\tfrac{1}{2}$Answer option
- CTwo electrons in $2p$ with $m_\ell=-1$, $m_s=+\tfrac{1}{2}$ and $m_\ell=+1$, $m_s=-\tfrac{1}{2}$Answer option
- DTwo electrons in $2p$ with $m_\ell=0$, $m_s=+\tfrac{1}{2}$ and $m_\ell=0$, $m_s=+\tfrac{1}{2}$Correct answer
Why this answer is right
Pauli forbids two electrons sharing all quantum numbers. Choice D has identical $n,\ell,m_\ell,m_s$ for both electrons, which is impossible.
An electron in an atom is specified by $n,\ell,m_\ell,m_s$. The Pauli principle states that no two electrons can share the same set of four quantum numbers.
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.