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Indian Navy INET Class9 Atom Structure — practice questions

Explore 38 free Class9 Atom Structure questions in Indian Navy INET Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Class9 Atom Structure questions and exercises with answers

Showing questions 1–30 of 38.

The minimum energy required to remove an electron from the ground state of a hydrogen atom is:Read the answer and worked solution →Electron configuration of nitrogen (Z = 7):Read the answer and worked solution →Number of nodes in a 3p orbital (n=3, l=1):Read the answer and worked solution →In the photoelectric effect, the kinetic energy of the ejected electron depends onRead the answer and worked solution →Rydberg's formula for the wave number of a hydrogen spectral line isRead the answer and worked solution →An electron is accelerated through a potential difference of $100\,V$. Its de Broglie wavelength is approximatelyRead the answer and worked solution →The ionisation energy of hydrogen in its ground state isRead the answer and worked solution →The Bohr radius for hydrogen ($n=1, Z=1$) is approximatelyRead the answer and worked solution →The proton was discovered through experiments involvingRead the answer and worked solution →As per your syllabus, eqn 2.15, the radius of the n-th Bohr orbit for a hydrogen-like ion (one electron, atomic number Z) scales as which expression?Read the answer and worked solution →As per your syllabus, in the hydrogen-atom spectrum the BALMER series corresponds to transitions ending in n₁ = 2 and lies in which spectral region?Read the answer and worked solution →In the photoelectric effect, the kinetic energy of ejected electrons is:Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →True or false: Bohr's modelRead the answer and worked solution →Complete the reason Bohr's model failsRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Match each quantum number to what it is calledRead the answer and worked solution →True or false: the quantum mechanical modelRead the answer and worked solution →Order the structure: from atom down to electronRead the answer and worked solution →Complete the (n + l) ruleRead the answer and worked solution →True or false: orbital energiesRead the answer and worked solution →Match each rule to what it forbids or requiresRead the answer and worked solution →Tap every statement Hund's rule requiresRead the answer and worked solution →The fundamental particlesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →

Worked practice questions

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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Electron configuration of nitrogen (Z = 7):

  1. 2s² 2p⁵
  2. 1s² 2s² 2p⁴ 3p¹
  3. 1s² 2s² 2p³
  4. 1s² 2s² 2p² 3s¹
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Correct answer: C. 1s² 2s² 2p³

**N has 7 electrons:** 1s² 2s² 2p³. The 2p³ has 3 unpaired electrons by Hund's rule (each in a separate p-orbital). Very stable half-filled state, contributing to N's special inertness.

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Number of nodes in a 3p orbital (n=3, l=1):

  1. 0
  2. 2 (n-1 total nodes; 1 angular + 1 radial)
  3. 3
  4. 1
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Correct answer: B. 2 (n-1 total nodes; 1 angular + 1 radial)

Total nodes = n - 1 = 2 for n=3. Angular nodes = l = 1 (the planar node through the nucleus). Radial nodes = n - l - 1 = 1. So 1 angular + 1 radial = 2 total nodes.

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More Chemistry topics

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