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JEE Main Structure of the Atom — practice questions

Explore 5 free Structure of the Atom questions in JEE Main Science. Open each question or exercise to read its content, reference answers and available explanation.

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Structure of the Atom questions and exercises with answers

Showing questions 1–5 of 5.

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 →

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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