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PUCET (Punjab) Atomic combinations — practice questions

Explore 103 free Atomic combinations questions in PUCET (Punjab) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Atomic combinations questions and exercises with answers

Showing questions 1–30 of 103.

The maximum number of electrons that can be accommodated in the $n$-th shell of an atom is:Read the answer and worked solution →The four quantum numbers describing an electron in an atom are:Read the answer and worked solution →According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \geq$:Read the answer and worked solution →Two electrons in the same orbital must have:Read the answer and worked solution →Calculate the de Broglie wavelength of an electron moving at $10^6$ m/s. (Take $h = 6.6 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg)Read the answer and worked solution →Bohr's first postulate: electrons revolve in ___ orbits around the nucleus.Read the answer and worked solution →In Bohr's model, angular momentum of an electron in orbit n equals:Read the answer and worked solution →Energy of electron in n-th orbit of hydrogen:Read the answer and worked solution →Principal quantum number n indicates:Read the answer and worked solution →Number of orbitals in subshell of azimuthal quantum number l:Read the answer and worked solution →How many electrons can fit in a single orbital?Read the answer and worked solution →Electron configuration of carbon (Z = 6):Read the answer and worked solution →For hydrogen, wavelength emitted in n=3 to n=2 transition (Balmer):Read the answer and worked solution →de Broglie wavelength of an electron with momentum p:Read the answer and worked solution →Heisenberg uncertainty principle:Read the answer and worked solution →Number of subshells in shell with n = 3:Read the answer and worked solution →Electron configuration of nitrogen (Z = 7):Read the answer and worked solution →Total number of electrons that can occupy shell with n = 4:Read the answer and worked solution →Magnetic quantum number m_l for a 3d orbital can be:Read the answer and worked solution →Wavelength of light needed to ionize hydrogen from ground state (13.6 eV):Read the answer and worked solution →Radius of n-th Bohr orbit for hydrogen (a₀ ≈ 0.529 Å):Read the answer and worked solution →Effective nuclear charge Z_eff for the 2p electron of carbon (using Slater's rules approximately):Read the answer and worked solution →For He⁺ ion (1 electron, Z = 2), energy in ground state:Read the answer and worked solution →The Pauli exclusion principle states that two electrons in an atom cannot have:Read the answer and worked solution →Energy of the photon emitted in a hydrogen transition from n=4 to n=2:Read the answer and worked solution →Number of nodes in a 3p orbital (n=3, l=1):Read the answer and worked solution →For Bohr's hydrogen-like atom, what is the velocity of electron in n-th orbit (in terms of fine structure constant α and c)?Read the answer and worked solution →Photoelectric effect: when light of frequency ν falls on a metal of work function φ, KE of emitted electrons is:Read the answer and worked solution →The wave number $\bar{\nu}$ of electromagnetic radiation is defined asRead the answer and worked solution →In vacuum, the velocity of electromagnetic radiation isRead the answer and worked solution →

Worked practice questions

The maximum number of electrons that can be accommodated in the $n$-th shell of an atom is:

  1. $n^2$
  2. $n$
  3. $2n$
  4. $2n^2$
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Correct answer: D. $2n^2$

Bohr-Bury rule. Each shell of principal quantum number $n$ contains $n^2$ orbitals, and each orbital holds at most 2 electrons (Pauli). So shell capacity is $2n^2$. Quick check: K shell ($n=1$) holds 2, L ($n=2$) holds 8, M ($n=3$) holds 18, N ($n=4$) holds 32. All powers-of-two-friendly numbers.

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The four quantum numbers describing an electron in an atom are:

  1. $n, l, m_l, m_s$
  2. $n, l, m_l, m_l$
  3. $n, p, d, f$
  4. $n, l, m, s$
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Correct answer: A. $n, l, m_l, m_s$

**Principal $n$** sets the shell and energy. **Azimuthal $l$** sets the subshell shape (s, p, d, f). **Magnetic $m_l$** sets the orbital's orientation in space. **Spin $m_s$** picks $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$. First three come from the Schrödinger equation the spin comes from electron's intrinsic angular momentum. Option D lists subshell labels (s, p, d, f), not quantum numbers.

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According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \geq$:

  1. $\dfrac{h}{2\pi}$
  2. $h$
  3. $\dfrac{h}{4\pi}$
  4. $\dfrac{h}{4}$
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Correct answer: C. $\dfrac{h}{4\pi}$

$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} = \dfrac{\hbar}{2}$, where $\hbar = h/2\pi$ is the reduced Planck constant. **Meaning:** you cannot simultaneously know an electron's position and momentum with arbitrary precision. Trying to pin one down makes the other less knowable. This forces us to talk about probability clouds (orbitals) rather than precise electron trajectories. Option B is often a near miss in memory because $\hbar = h/2\pi$, but the uncertainty bound has the factor of 2 baked into the denominator twice over.

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