Atomic spectra lab answers for A Level physics (free PDF download)

Question 1 Averaging both sides

If the grating is not exactly centred or the slit is not on the middle mark, one side reads too large and the other too small by the same amount. Averaging cancels this. The mean x is a better value than either single reading.

Question 2 Grating spacing

d = 1 / 600 mm = 1.667 × 10⁻³ mm = 1.67 × 10⁻⁶ m (1667 nm).

Question 3 Calculating wavelengths

For each line tan θ = x / L, then λ = d sin θ for the first order.
red: tan θ = 42.80 / 100.0 = 0.4280, θ = 23.17°, sin θ = 0.3935, λ = 1667 × 0.3935 = 656 nm.
blue-green: tan θ = 30.55 / 100.0 = 0.3055, θ = 16.99°, sin θ = 0.2922, λ = 1667 × 0.2922 = 487 nm.
violet: tan θ = 27.00 / 100.0 = 0.2700, θ = 15.11°, sin θ = 0.2607, λ = 1667 × 0.2607 = 434 nm.

Question 4 Comparing with accepted values

Percentage difference = |measured − accepted| / accepted × 100 %.
red: |656 − 656| / 656 = 0 %.
blue-green: |487 − 486| / 486 = 0.21 %.
violet: |434 − 434| / 434 = 0 %.
All three agree within about 0.5 %, which is good for a school grating.

Question 5 Identifying transitions

Photon energy = 1240 / λ.
red: 1.89 eV
blue-green: 2.55 eV
violet: 2.86 eV
These match the hydrogen gaps 3 → 2 (1.89 eV), 4 → 2 (2.55 eV) and 5 → 2 (2.86 eV). Red is 3 → 2, blue-green is 4 → 2, violet is 5 → 2; all end on n = 2.

Question 6 Uncertainty in the red line

Percentage uncertainty in x = 0.1 / 42.8 × 100 % = 0.23 %. Because λ = d sin θ and sin θ increases with x, a small fractional change in x gives a slightly smaller fractional change in λ (about 0.2 %, roughly ±1 nm). Using a longer L (such as 2 m) makes x larger, so the same ±0.1 cm is a smaller fraction.

Question 7 Second order

For the second order sin θ = 2λ / d = 2 × 656 / 1667 = 0.787, so θ = 51.9° and x = L tan θ = 128 cm. This is beyond the end of a one-metre scale, and the line is also dimmer, so it is hard to see.