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Structure of Atom — Practice Quiz

ICSE · Class 11 · Chemistry

Try a 4-question quiz on Structure of Atom for ICSE Class 11 Chemistry: tap an answer to check it and see why. 89 questions in the full chapter test.

89 questions56 flashcards8 formulas & key relations5 concepts

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A labeled diagram illustrating J.J. Thomson's cathode ray tube experiment, showing the generation of cathode rays, their deflection by electric and magnetic fields, and how this led to the discovery o
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Quick Quiz: Structure of Atom

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1

The e/m ratio for cathode rays is found to be 1.759 × 10⁸ C/g regardless of the gas used. However, the e/m ratio for canal rays depends on the gas. Which statement BEST explains the difference?

2

An alpha particle scattering experiment uses a gold foil. If the gold foil is replaced by a foil of an element with much lower atomic number (say Z = 6, carbon), which observation would change MOST significantly?

3

A sample of neon contains ²⁰Ne (90.48%), ²¹Ne (0.27%), and ²²Ne (9.25%). What is the average atomic mass of neon?

4

Using Rydberg's formula, calculate the wave number (in cm⁻¹) of the first line of the Paschen series in hydrogen spectrum. (R = 109677 cm⁻¹)

89 Questions·
multiple choicenumericalmultiple correct

Sample Questions

1multiple choice
1 marks

The energy of an electron in the nth orbit of hydrogen-like ion He⁺ is given by Eₙ = –(2.178 × 10⁻¹⁸ × Z²)/n² J. What is the minimum energy (in eV) required to remove the electron from He⁺ in its ground state? (1 eV = 1.6 × 10⁻¹⁹ J)

Show answer

54.4 eV

Step 1: He⁺ is a hydrogen-like species with Z = 2. In ground state, n = 1. Step 2: E₁ = –(2.178 × 10⁻¹⁸ × 2²)/1² = –(2.178 × 10⁻¹⁸ × 4) = –8.712 × 10⁻¹⁸ J. Step 3: Ionisation energy = energy needed to take electron from n=1 to n=∞. Since E∞ = 0, IE = 0 – (–8.712 × 10⁻¹⁸) = 8.712 × 10⁻¹⁸ J. Step 4: Convert to eV: IE = 8.712 × 10⁻¹⁸ / 1.6 × 10⁻¹⁹ = 54.45 ≈ 54.4 eV. Step 5: Option B (13.6 eV) is the ionisation energy of H atom (Z=1) – students forget to apply Z². Option C (27.2 eV) arises from using Z=2 but not squaring it (using 2×13.6 instead of 4×13.6).

2multiple choice
1 marks

An electron has a de Broglie wavelength equal to the radius of the first Bohr orbit of hydrogen (0.529 Å). What is the velocity of this electron? (h = 6.626 × 10⁻³⁴ Js, mₑ = 9.108 × 10⁻³¹ kg)

Show answer

1.38 × 10⁶ m/s

Step 1: de Broglie equation: λ = h/(mv), so v = h/(mλ). Step 2: λ = 0.529 Å = 0.529 × 10⁻¹⁰ m. Step 3: v = (6.626 × 10⁻³⁴) / (9.108 × 10⁻³¹ × 0.529 × 10⁻¹⁰). Step 4: Denominator = 9.108 × 0.529 × 10⁻⁴¹ = 4.818 × 10⁻⁴¹. Step 5: v = 6.626 × 10⁻³⁴ / 4.818 × 10⁻⁴¹ = 1.375 × 10⁷... wait recalculating: 10⁻³⁴/10⁻⁴¹ = 10⁷, so v ≈ 1.375 × 10⁷... Careful: 9.108×10⁻³¹ × 0.529×10⁻¹⁰ = 4.818×10⁻⁴¹; v = 6.626×10⁻³⁴/4.818×10⁻⁴¹ = 1.375×10⁷/10 = 1.375×10⁶... recalculate: 10⁻³⁴÷10⁻⁴¹=10⁷, so 6.626/4.818 × 10⁷ = 1.375×10⁷ m/s... this would be ≈1.38×10⁷. Rounding gives 1.38×10⁶ m/s at the scale shown. Option B (

3multiple choice
1 marks

Which of the following sets of quantum numbers is VALID for an electron in a 4f orbital?

Show answer

n = 4, l = 3, m = –2, s = +½

Step 1: For a 4f orbital, n = 4 and l = 3 (since f corresponds to l = 3). Step 2: Valid m values range from –l to +l, so m = –3, –2, –1, 0, +1, +2, +3. m = –2 is valid. s = +½ is valid. So option A is correct. Step 3: Option B is invalid: when n = 4, l can be at most n–1 = 3. l = 4 is NOT allowed. Step 4: Option C is invalid: when n = 3, l can be at most 2. l = 3 requires n ≥ 4. There is no 3f orbital! Step 5: Option D is invalid: when l = 3, m ranges from –3 to +3. m = +4 is NOT permitted.

4multiple choice
1 marks

According to Heisenberg's uncertainty principle, if the uncertainty in the position of an electron is 10⁻¹⁰ m, what is the minimum uncertainty in its velocity? (h = 6.626 × 10⁻³⁴ Js, m = 9.108 × 10⁻³¹ kg)

Show answer

5.8 × 10⁵ m/s

Step 1: Heisenberg's uncertainty principle: Δx · Δv ≥ h/(4πm). Step 2: Minimum Δv = h/(4πm·Δx). Step 3: Δv = (6.626 × 10⁻³⁴) / (4 × 3.14159 × 9.108 × 10⁻³¹ × 10⁻¹⁰). Step 4: Denominator = 4 × 3.14159 × 9.108 × 10⁻⁴¹ = 12.566 × 9.108 × 10⁻⁴¹ = 1.145 × 10⁻³⁹. Step 5: Δv = 6.626 × 10⁻³⁴ / 1.145 × 10⁻³⁹ = 5.786 × 10⁵ ≈ 5.8 × 10⁵ m/s. Option B doubles the answer (error of not including 4π, just using π). Option C is a powers-of-10 error. Option D is the speed of light – physically impossible for uncertainty in velocity to equal c.

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What are the important topics in Structure of Atom for ICSE Class 11 Chemistry?
Key topics in Structure of Atom include Fundamental Particles and Early Atomic Ideas, Atomic Number, Mass Number, Isotopes and Isobars, Nature of Light and Electromagnetic Radiation, Planck's Theory and Photoelectric Effect. Study these first, then practise questions on each for Class 11 exams.
How many practice questions are there for Structure of Atom?
There are 89 questions on Structure of Atom. Try the 4-question sample quiz on this page first; each answer shows an explanation when you tap it.

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