1. (a) Explain two uses of mass spectrometer

1. Determine isotopic masses & abundances – separates isotopes and measures their relative abundances to obtain the relative atomic mass.
2. Identify unknown compounds – analytical tool in forensics, pharmaceuticals and space science to identify molecules and detect impurities.

1. (b) One student observed radiation from an electron in a discharge tube falling from energy level n=2 to the ground level n=1.

(i) Type of spectrum observed – Line emission spectrum (discrete lines due to quantized energy differences).

(ii) Frequency of this radiation

Energy difference:
ΔE = 13.6 eV (1/1² - 1/2²)
ΔE = 13.6 (1 - 0.25) = 10.2 eV

Convert to joules:
E = 10.2 × 1.6 × 10⁻¹⁹ = 1.632 × 10⁻¹⁸ J

Frequency:
f = E / h = 1.632 × 10⁻¹⁸ / 6.63 × 10⁻³⁴
f = 2.46 × 10¹⁵ Hz
        

1. (c) Was there a need for Heisenberg to challenge Bohr’s assumption of definite position and velocity?

Yes. Heisenberg’s Uncertainty Principle states that position and momentum cannot both be known exactly at the same time. Bohr’s model assumes both are definite, so Heisenberg’s challenge led to the quantum-mechanical model of the atom.

2. (a) Which statements in Dalton’s atomic theory are now incorrect? (Give three and why.)

1. All atoms of an element are identical. Incorrect: isotopes of the same element have different masses.
2. Atoms are indivisible. Incorrect: atoms contain subatomic particles (protons, neutrons, electrons).
3. Atoms cannot be created or destroyed. Incorrect: nuclear reactions create/destroy atoms.

2. (b) The wavelength of a certain electromagnetic radiation is 1.20 cm. Find photon energy.

λ = 1.20 cm = 0.012 m

Frequency:
f = c / λ = 3.0 × 10⁸ / 0.012
f = 2.5 × 10¹⁰ Hz

Energy:
E = h f = 6.63 × 10⁻³⁴ × 2.5 × 10¹⁰
E = 1.66 × 10⁻²³ J
        

2. (c) Natural Cu has Ar = 63.55. Its isotopes are 63Cu (62.9) and 65Cu (64.9). Find % abundance.

Let % of 63Cu = x, % of 65Cu = 100 - x

(62.9x + 64.9(100 - x)) / 100 = 63.55

62.9x + 6490 - 64.9x = 6355
-2x = -135
x = 67.5 %

So:
63Cu = 67.5 %
65Cu = 32.5 %
        

3. (a) Differences

• Subsidiary quantum number (ℓ): determines orbital shape (s, p, d, f).
• Magnetic quantum number (mℓ): determines orbital orientation in space.
• Orbitals: regions of high probability of finding an electron.
• Degenerate orbitals: orbitals with the same energy but different orientations (e.g., px, py, pz).

3. (b) (i) For principal quantum number n = 2:

ℓ = 0 → mℓ = 0 → 2 electrons (2s orbital)
ℓ = 1 → mℓ = -1,0,+1 → 6 electrons (2p orbitals)

Total electrons in n=2 level = 8
        

3. (b) (ii) Compare de Broglie wavelengths.

A ball of 0.21 kg at 30 m/s has a negligible wavelength due to large mass. An electron at the same speed has a measurable wavelength. Thus wave behavior is observable for electrons but negligible for macroscopic objects.

4. (a) Rutherford scattering led to:

1. Positive charge concentrated in the nucleus.
2. Most of the atom is empty space.

4. (b) Which set has higher energy?

A. n=3, ℓ=2, m=+1  (3d orbital)
B. n=3, ℓ=1, m=-1  (3p orbital)

Answer: A has higher energy.
        

4. (c) How many electrons may have the following quantum numbers?

(i) n=4, ms=+1/2 → 16 electrons
(ii) n=4, ℓ=0 → 2 electrons
(iii) n=3, ℓ=1, mℓ=-1 → 2 electrons
        

4. (d) Wavelength of an electron with KE = 4.55 × 10⁻²⁵ J

p = √(2 m E)
  = √(2 × 9.11 × 10⁻³¹ × 4.55 × 10⁻²⁵)
  = 9.1 × 10⁻²⁸ kg m/s

λ = h / p
  = 6.63 × 10⁻³⁴ / 9.1 × 10⁻²⁸
  = 7.3 × 10⁻⁷ m