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Indian Navy AA (Artificer Apprentice) Atomic Structure — practice questions

Explore 33 free Atomic Structure questions in Indian Navy AA (Artificer Apprentice) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Atomic Structure questions and exercises with answers

Showing questions 1–30 of 33.

The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:Read the answer and worked solution →A photon of wavelength $300\,nm$ has approximately what energy in eV? (Use $hc = 1240\,eV\,nm$.)Read the answer and worked solution →In the photoelectric effect, the kinetic energy of the ejected electron depends onRead the answer and worked solution →The number of *radial* nodes in a $4s$ orbital isRead the answer and worked solution →For $He^+$, the ionisation energy from its ground state isRead the answer and worked solution →Which of the following is **NOT** a possible set of quantum numbers for a 3d electron?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 →As per your syllabus, for n = 3 the TOTAL number of orbitals is which value?Read the answer and worked solution →A 3p orbital is specified by:Read the answer and worked solution →Match each scientist to their contributionRead the answer and worked solution →Ionic or covalent?Read the answer and worked solution →Complete the driver of bond formationRead the answer and worked solution →Single, double and triple bondsRead the answer and worked solution →True or false: Lewis structuresRead the answer and worked solution →Complete the definition of resonanceRead the answer and worked solution →Tap every bond parameter the chapter namesRead the answer and worked solution →True or false: ionic solidsRead 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 →Bonded pair or lone pair?Read the answer and worked solution →Flip and pair: term with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Order the reasoning: why resonance is neededRead the answer and worked solution →Reading the definitions backRead the answer and worked solution →True or false: what bond parameters affectRead the answer and worked solution →Bond parameters in useRead the answer and worked solution →Which parameter answers this?Read the answer and worked solution →True or false: resonanceRead the answer and worked solution →Order the steps: draw a Lewis structureRead the answer and worked solution →

Worked practice questions

The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:

  1. Lyman series
  2. Paschen series
  3. Pfund series
  4. Balmer series
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Correct answer: D. Balmer series

Balmer series corresponds to transitions terminating in $n_1 = 2$. The first four lines ($n_2 = 3, 4, 5, 6$ to $n_1 = 2$) lie at approximately $656\,\text{nm}$ (H$_\alpha$, red), $486\,\text{nm}$ (H$_\beta$, blue-green), $434\,\text{nm}$ (H$_\gamma$, blue) and $410\,\text{nm}$ (H$_\delta$, violet), all in the visible range.

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A photon of wavelength $300\,nm$ has approximately what energy in eV? (Use $hc = 1240\,eV\,nm$.)

  1. $2.07\,eV$
  2. $8.26\,eV$
  3. $4.13\,eV$
  4. $0.413\,eV$
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Correct answer: C. $4.13\,eV$

$E = hc/\lambda = 1240/300 \approx 4.13\,eV$. The shortcut $hc = 1240\,eV\,nm$ is worth memorising — divides wavelength in nm directly to give energy in eV.

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In the photoelectric effect, the kinetic energy of the ejected electron depends on

  1. the intensity of the incident light
  2. the frequency of the incident light
  3. the colour of the metal surface
  4. the temperature of the metal
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Correct answer: B. the frequency of the incident light

$KE_{max} = h\nu - \phi$ where $\phi$ is the work function. KE depends only on *frequency* intensity only changes the *number* of ejected electrons, not their energy. This is Einstein's resolution of the photoelectric puzzle.

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