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Rajasthan Board Class 12 (RBSE) Atomic Structure and Valency — practice questions

Explore 8 free Atomic Structure and Valency questions in Rajasthan Board Class 12 (RBSE) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Atomic Structure and Valency questions and exercises with answers

Showing questions 1–8 of 8.

The radius of the $n$-th Bohr orbit of a hydrogen-like atom of atomic number $Z$ varies as:Read the answer and worked solution →The minimum energy required to remove an electron from the ground state of a hydrogen atom is:Read the answer and worked solution →The kinetic energy of an electron in the $n$-th Bohr orbit of a hydrogen atom is related to its total energy $E_n$ by:Read the answer and worked solution →Which spectral series of hydrogen lies entirely in the ultraviolet region?Read the answer and worked solution →The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:Read the answer and worked solution →The major shortcoming of the Rutherford atomic model was that it could not explain:Read the answer and worked solution →For a hydrogen-like ion of atomic number $Z$, the ground-state energy is:Read the answer and worked solution →An electron in a hydrogen atom undergoes a transition from $n = 3$ to $n = 2$. The energy of the emitted photon is approximately:Read the answer and worked solution →

Worked practice questions

The radius of the $n$-th Bohr orbit of a hydrogen-like atom of atomic number $Z$ varies as:

  1. $r \propto n / Z^2$
  2. $r \propto Z / n^2$
  3. $r \propto n^2 / Z$
  4. $r \propto Z n^2$
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Correct answer: C. $r \propto n^2 / Z$

From Bohr's derivation, $r_n = \dfrac{n^2 h^2}{4\pi^2 m k Z e^2}$, hence $r_n \propto n^2 / Z$. **Therefore higher orbits are larger ($n^2$) and higher-$Z$ atoms hold their electrons in tighter orbits ($1/Z$).**

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The minimum energy required to remove an electron from the ground state of a hydrogen atom is:

  1. $3.4\,\text{eV}$
  2. $13.6\,\text{eV}$
  3. $27.2\,\text{eV}$
  4. $1.51\,\text{eV}$
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Correct answer: B. $13.6\,\text{eV}$

Ionisation energy is the energy required to take the electron from $n = 1$ to $n = \infty$. $E_\infty - E_1 = 0 - (-13.6) = 13.6\,\text{eV}$. The value $27.2\,\text{eV}$ is the magnitude of the potential energy in the ground state, not the ionisation energy.

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The kinetic energy of an electron in the $n$-th Bohr orbit of a hydrogen atom is related to its total energy $E_n$ by:

  1. $E_k = E_n$
  2. $E_k = 2E_n$
  3. $E_k = -E_n$
  4. $E_k = -E_n / 2$
Show answer and solution

Correct answer: C. $E_k = -E_n$

For the Bohr atom the magnitudes of kinetic and total energy are equal but opposite in sign: $E_k = +13.6 Z^2/n^2\,\text{eV}$ and $E_n = -13.6 Z^2/n^2\,\text{eV}$. Hence $E_k = -E_n$. Equivalently $E_k = -E_n$ and $U = 2E_n$, with $U + E_k = E_n$.

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More Chemistry topics

Coordination CompoundsNamed ReactionsAlcohols Phenols EthersChemical KineticsSolid StateAlkenesAtomic combinationsAtoms and Elements