DP Physics (last assessment 2024)

Question 18N.2.HL.TZ0.5

Date	November 2018	Marks available	[Maximum mark: 14]	Reference code	18N.2.HL.TZ0.5
Level	HL	Paper	2	Time zone	TZ0
Command term	Calculate, Determine, Explain, Identify, Outline, Show that	Question number	5	Adapted from	N/A

[Maximum mark: 14]

18N.2.HL.TZ0.5

The diagram shows the position of the principal lines in the visible spectrum of atomic hydrogen and some of the corresponding energy levels of the hydrogen atom.

(a.i)

Determine the energy of a photon of blue light (435nm) emitted in the hydrogen spectrum.

[3]

Markscheme

identifies λ = 435 nm ✔

E = « $\frac{{hc}}{\lambda }$ =» $\frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.35 \times {{10}^{ - 7}}}}$ ✔

4.6 ×10⁻¹⁹ «J» ✔

(a.ii)

Identify, with an arrow labelled B on the diagram, the transition in the hydrogen spectrum that gives rise to the photon with the energy in (a)(i).

[1]

Markscheme

–0.605 OR –0.870 OR –1.36 to –5.44 AND arrow pointing downwards ✔

Arrow MUST match calculation in (a)(i)

Allow ECF from (a)(i)

(a.iii)

Explain your answer to (a)(ii).

[2]

Markscheme

Difference in energy levels is equal to the energy of the photon ✔

Downward arrow as energy is lost by hydrogen/energy is given out in the photon/the electron falls from a higher energy level to a lower one ✔

A low-pressure hydrogen discharge lamp contains a small amount of deuterium gas in addition to the hydrogen gas. The deuterium spectrum contains a red line with a wavelength very close to that of the hydrogen red line. The wavelengths for the principal lines in the visible spectra of deuterium and hydrogen are given in the table.

Light from the discharge lamp is normally incident on a diffraction grating.

(b.i)

The light illuminates a width of 3.5 mm of the grating. The deuterium and hydrogen red lines can just be resolved in the second-order spectrum of the diffraction grating. Show that the grating spacing of the diffraction grating is about 2 × 10^–6m.

[2]

Markscheme

$\frac{\lambda }{{2\Delta \lambda }} = \frac{{656.20}}{{0.181 \times 2}} = 1813$ «lines» ✔

so spacing is $\frac{{3.5 \times {{10}^{ - 3}}}}{{1813}}$ «= 1.9 × 10⁻⁶ m» ✔

Allow use of either wavelength or the mean value

Must see at least 2 SF for a bald correct answer

(b.ii)

Calculate the angle between the first-order line of the red light in the hydrogen spectrum and the second-order line of the violet light in the hydrogen spectrum.

[3]

Markscheme

2 × 4.1 × 10⁻⁷ = 1.9 × 10⁻⁶ sin θ_v seen

6.6 × 10⁻⁷ = 1.9 × 10⁻⁶ sin θ_r seen ✔

θ_v = 24 − 26 «°»

θ_r = 19 − 20 «°» ✔

Δθ = 5 − 6 «°» ✔

For MP3 answer must follow from answers in MP2

For MP3 do not allow ECF from incorrect angles

(b.iii)

The light source is changed so that white light is incident on the diffraction grating. Outline the appearance of the diffraction pattern formed with white light.

[3]

Markscheme

centre of pattern is white ✔

coloured fringes are formed ✔

blue/violet edge of order is closer to centre of pattern

red edge of order is furthest from centre of pattern ✔

the greater the order the wider the pattern ✔

there are gaps between «first and second order» spectra ✔