# IBDP Physics 12.1 – The interaction of matter with radiation: IB Style Question Bank HL Paper 2

### Question

The de Broglie wavelength λ of a particle accelerated close to the speed of light is approximately

λ = $\frac{hc}{E}$

where E is the energy of the particle.

A beam of electrons of energy 4.2 × 108 eV is produced in an accelerator.

(a) Show that the wavelength of an electron in the beam is about 3 × 1015 m.                     [1]

Ans

$$\lambda =\frac{6.63\times10^{-34}\times3\times10^8}{1.60\times10^{-19}\times4.2\times10^8} = 2.96\times10^{-15}m$$

Question

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the ground state. Photons are emitted and are incident on a photoelectric surface as shown.

The photons cause the emission of electrons from the photoelectric surface. The work function of the photoelectric surface is 5.1 eV.

The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so that the collecting plate is at –1.2 V.

a.

Show that the energy of photons from the UV lamp is about 10 eV.

[2]
b.i.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

[2]
b.ii.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

[2]
b.iii.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

[1]
c.i.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

[2]
c.ii.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

[2]

## Markscheme

a.

E1 = –13.6 «eV» E2 = – $$\frac{{13.6}}{4}$$ = –3.4 «eV»

energy of photon is difference E2E1 = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

b.i.

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10–19 = 7.8 × 10–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10−19 «J».

b.ii.

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

b.iii.

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

c.i.

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

c.ii.

kinetic energy at collecting plate = 0.9 «eV»

speed = «$$\sqrt {\frac{{2 \times 0.9 \times 1.6 \times {{10}^{ – 19}}}}{{9.11 \times {{10}^{ – 31}}}}}$$» = 5.6 × 105 «ms–1»

Allow ECF from MP1

[2 marks]

Question

Yellow light of photon energy 3.5 x 10–19 J is incident on the surface of a particular photocell.

The photocell is connected to a cell as shown. The photoelectric current is at its maximum value (the saturation current).

Radiation with a greater photon energy than that in (b) is now incident on the photocell. The intensity of this radiation is the same as that in (b).

a.i.

Calculate the wavelength of the light.

[1]
a.ii.

Electrons emitted from the surface of the photocell have almost no kinetic energy. Explain why this does not contradict the law of conservation of energy.

[2]
b.

Radiation of photon energy 5.2 x 10–19 J is now incident on the photocell. Calculate the maximum velocity of the emitted electrons.

[2]
c.i.

Describe the change in the number of photons per second incident on the surface of the photocell.

[1]
c.ii.

State and explain the effect on the maximum photoelectric current as a result of increasing the photon energy in this way.

[3]

## Markscheme

a.i.

wavelength = «$$\frac{{hc}}{E} = \frac{{1.99 \times {{10}^{ – 25}}}}{{{\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ – 19}}}} =$$» 5.7 x 10–7 «m»

If no unit assume m.

a.ii.

«potential» energy is required to leave surface

Do not allow reference to “binding energy”.
Ignore statements of conservation of energy.

all/most energy given to potential «so none left for kinetic energy»

b.

energy surplus = 1.7 x 10–19 J

vmax = $$\sqrt {\frac{{2 \times 1.7 \times {{10}^{ – 19}}}}{{9.1 \times {{10}^{ – 31}}}}} = 6.1 \times {10^5}$$ «m s–1»

Award [1 max] if surplus of 5.2 x 10–19J used (answer: 1.1 x 106 m s–1)

c.i.

«same intensity of radiation so same total energy delivered per square metre per second»

light has higher photon energy so fewer photons incident per second

Reason is required

c.ii.

1:1 correspondence between photon and electron

so fewer electrons per second

current smaller

Allow ECF from (c)(i)
Allow ECF from MP2 to MP3.