CBSE Class 11 Chemistry Structure of Atom Worksheet Set 09

Structure of Atom Worksheet

Question 1. State the postulates of Bohr's model of the hydrogen atom.
Answer:
1. The electron moves round the nucleus in certain fixed circular paths called orbits or stationary states, each with a fixed energy.
2. While an electron stays in one orbit it neither gains nor loses energy.
3. Energy is absorbed or emitted only when the electron jumps from one orbit to another: ΔE = E₂ − E₁ = hν.
4. The angular momentum of the electron is quantised: mvr = nh/2π, where n = 1, 2, 3 ...
Results. E_n = −2.18 × 10⁻¹⁸ × Z²/n² J per atom, and r_n = 52.9 × n²/Z pm.
In simple words: Only certain orbits are allowed, and light is emitted or absorbed when the electron jumps between them.

Teacher's Note:
1. Quantised angular momentum picks out the allowed orbits.
2. The model applies to H and one-electron ions such as He⁺ and Li²⁺.

Question 2. The energy of the electron in the first orbit of hydrogen is −2.18 × 10⁻¹⁸ J per atom. Calculate the energy of the fifth orbit and the radius of the fifth Bohr orbit (r_n = 0.529 n² Å).
Answer:
E₅ = E₁/n² = −2.18 × 10⁻¹⁸ ÷ 25 = −8.72 × 10⁻²⁰ J.
r₅ = 0.529 × 25 = 13.23 Å (1.32 nm).
In simple words: Energy falls as 1/n² and the radius grows as n².

Teacher's Note:
1. The fifth orbit is much higher and larger than the first.
2. 1 Å = 0.1 nm.

Question 3. Calculate the wavelength and energy of the radiation emitted when an electron of a hydrogen atom falls from n = 4 to n = 2. Name the series and the region.
Answer:
ΔE = 2.18 × 10⁻¹⁸ (1/4 − 1/16) = 4.09 × 10⁻¹⁹ J.
λ = hc/ΔE = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ 4.09 × 10⁻¹⁹ = 4.86 × 10⁻⁷ m (486 nm).
A fall to n = 2 belongs to the Balmer series, in the visible region (blue-green line).
In simple words: Find the energy gap, then convert it to a wavelength with λ = hc/ΔE.

Teacher's Note:
1. Balmer lines all end at n = 2.
2. The Rydberg formula gives the same answer: 1/λ = 1.097 × 10⁷ × 3/16.

Question 4. Which transition in the hydrogen spectrum has the same wavelength as the n = 4 to n = 2 transition of He⁺?
Answer:
For a one-electron species, 1/λ = R Z² (1/n₁² − 1/n₂²).
He⁺: Z = 2, so 1/λ = R × 4 × (1/4 − 1/16) = R × 3/4.
Hydrogen: 1/λ = R (1/n₁² − 1/n₂²). For the wavelengths to be equal, 1/n₁² − 1/n₂² = 3/4, which fits n₁ = 1 and n₂ = 2.
So the n = 2 to n = 1 transition of hydrogen matches.
In simple words: He⁺ levels are four times as far apart as in H, so a higher pair of levels in He⁺ matches a lower pair in H.

Teacher's Note:
1. Equate Z²(1/n₁² − 1/n₂²) for the two species.
2. Test small whole numbers for n₁ and n₂.

Question 5. Calculate the energy required to remove the electron from (a) He⁺ and (b) Li²⁺ in their ground states. (Ionisation energy of H is 2.18 × 10⁻¹⁸ J, or 13.6 eV.)
Answer:
For a one-electron ion in n = 1, I.E. = I.E.(H) × Z².
(a) He⁺ (Z = 2): 2.18 × 10⁻¹⁸ × 4 = 8.72 × 10⁻¹⁸ J (54.4 eV).
(b) Li²⁺ (Z = 3): 2.18 × 10⁻¹⁸ × 9 = 1.96 × 10⁻¹⁷ J (122.4 eV).
In simple words: The nucleus pulls harder as Z grows, so the energy needed grows as Z².

Teacher's Note:
1. Take n₂ = ∞ for ionisation.
2. 13.6 eV × Z² gives the answer in electron volts.

Question 6. The wavelength of the first line of the Balmer series of hydrogen is 6561 Å. Calculate the wavelength of the second line and of the limiting line of the series.
Answer:
First line: n = 3 to 2: 1/λ₁ = R(1/4 − 1/9) = R × 5/36.
Second line: n = 4 to 2: 1/λ₂ = R(1/4 − 1/16) = R × 3/16.
λ₂ = λ₁ × (5/36) ÷ (3/16) = 6561 × 80/108 = 4860 Å.
Limiting line: n = ∞ to 2: 1/λ = R/4, so λ = 6561 × (5/36) × 4 = 3645 Å.
In simple words: Compare the Rydberg expressions for the lines instead of using the value of R.

Teacher's Note:
1. The limiting line has the shortest wavelength of the series.
2. Balmer lines get closer together towards the limit.

Teacher's Note:
1. Only the Balmer series is in the visible range.
2. Lyman is the highest-energy series.

Question 8. State three limitations of Bohr's model of the atom.
Answer:
1. It cannot explain the spectra of atoms with more than one electron, nor the finer details (fine structure) of the hydrogen spectrum.
2. It cannot explain the splitting of spectral lines in a magnetic field (Zeeman effect) or in an electric field (Stark effect).
3. It ignores the dual (wave and particle) nature of the electron and treats the orbit as a fixed path, which violates Heisenberg's uncertainty principle.
4. It cannot explain how atoms form molecules by chemical bonds.
In simple words: Bohr's model works for hydrogen but not for bigger atoms, and its fixed orbits are not allowed by quantum ideas.

Teacher's Note:
1. The uncertainty principle forbids exact paths for electrons.
2. A better model needed wave mechanics, i.e. the quantum mechanical model.

Question 9. Calculate the de Broglie wavelength of (a) an electron moving at 1% of the speed of light and (b) a golf ball of mass 40 g moving at 45 m s⁻¹. Why is the wave nature of the golf ball not seen?
Answer:
(a) v = 0.01 × 3 × 10⁸ = 3 × 10⁶ m s⁻¹. λ = h/mv = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 3 × 10⁶) = 2.42 × 10⁻¹⁰ m, comparable to atomic size.
(b) λ = 6.626 × 10⁻³⁴ ÷ (0.040 × 45) = 3.68 × 10⁻³⁴ m.
This is far smaller than any size we can detect, because the mass is so large.
In simple words: Wave behaviour shows up for electrons but is completely hidden for large objects.

Teacher's Note:
1. Convert grams to kilograms.
2. λ is inversely proportional to mass.

Question 10. Show that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of the de Broglie wavelength of the electron moving in it.
Answer:
Bohr's postulate: mvr = nh/2π, so 2πr = nh/mv. (1)
de Broglie: λ = h/mv. (2)
Putting (2) in (1): 2πr = nλ, with n = 1, 2, 3 ...
The circumference of the orbit is therefore a whole number of wavelengths, which means the electron wave fits the orbit exactly as a standing wave.
In simple words: Only orbits that hold a whole number of electron wavelengths are allowed.

Teacher's Note:
1. This links Bohr's quantisation with de Broglie's wave idea.
2. n is the number of wavelengths round the orbit.

Question 11. State Heisenberg's uncertainty principle. Calculate the uncertainty in the position of an electron whose speed is uncertain by 5.7 × 10⁵ m s⁻¹.
Answer:
Statement. It is impossible to determine the exact position and exact momentum (or velocity) of a microscopic particle at the same time. Δx · Δp ≥ h/4π.
Δx = h/(4π m Δv) = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 9.11 × 10⁻³¹ × 5.7 × 10⁵) = 1.0 × 10⁻¹⁰ m.
In simple words: Pinning down where an electron is makes its speed fuzzy, and the reverse is also true.

Teacher's Note:
1. This is about 1 Å, the size of an atom.
2. The principle rules out fixed electron orbits.

Question 12. A golf ball of mass 40 g moves at 45 m s⁻¹, with the speed measurable to within 2%. Find the minimum uncertainty in its position. Why can an electron not exist inside the nucleus (radius about 10⁻¹⁵ m)?
Answer:
Δv = 2% of 45 = 0.9 m s⁻¹. Δx = h/(4π m Δv) = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 0.040 × 0.9) = 1.46 × 10⁻³³ m, utterly negligible.
For an electron confined within 10⁻¹⁵ m, Δv = h/(4π m Δx) = 5.8 × 10¹⁰ m s⁻¹.
This is greater than the speed of light (3 × 10⁸ m s⁻¹), which is impossible, so an electron cannot exist inside the nucleus.
In simple words: The principle matters for electrons but not for everyday objects.

Teacher's Note:
1. Percentage of 45 m s⁻¹ gives Δv.
2. Compare the answer with the speed of light to judge if it is possible.

Question 13. The kinetic energy of an electron is 4.55 × 10⁻²⁵ J. Calculate its de Broglie wavelength.
Answer:
K.E. = ½mv², so v = √(2 K.E./m) = √(2 × 4.55 × 10⁻²⁵ ÷ 9.109 × 10⁻³¹) = 1.00 × 10³ m s⁻¹.
λ = h/mv = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 1.00 × 10³) = 7.28 × 10⁻⁷ m.
In simple words: First find the speed from the kinetic energy, then use h/mv.

Teacher's Note:
1. Alternatively λ = h/√(2m K.E.).
2. The speed here is only about 10³ m s⁻¹.