5.2 Interactions of electromagnetic radiation with matter — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Section 5.1.

Which group of the spectrum lies between infrared and ultraviolet?
Use $v = f\lambda$ to find $f$ when $v = 3.0 \times 10^8\,\text{m/s}$ and $\lambda = 0.50\,\text{m}$ .
Write 10 μm in metres.

Answers

Visible light.
$f = \dfrac{v}{\lambda} = \dfrac{3.0 \times 10^8}{0.50} = 6.0 \times 10^8\,\text{Hz}$ .
$1.0 \times 10^{-5}\,\text{m}$ .

If question 2 went wrong, look again at rearranging the wave equation. If question 3 went wrong, remember that micro means $10^{-6}$ .

Practise with help

1. Ultraviolet radiation has a frequency of $1.5 \times 10^{15}\,\text{Hz}$ . Calculate its wavelength.

Hint

This is a wave equation calculation. You know the speed and the frequency, and you want the wavelength.

Next step

Write $\lambda = \dfrac{c}{f}$ and put in $f = 1.5 \times 10^{15}\,\text{Hz}$ .

Full working

\begin{aligned} \lambda &= \frac{c}{f} \\ &= \frac{3.0 \times 10^8}{1.5 \times 10^{15}} \\ &= 2.0 \times 10^{-7}\,\text{m} \end{aligned}

This is 200 nm, which is in the ultraviolet range.

2. A black metal plate and a shiny metal plate are placed in bright sunshine. Explain which plate warms up faster, and why.

Hint

The radiation that reaches both plates is the same. The difference is what each surface does with it.

Next step

Say which surface absorbs more infrared radiation, and what happens to that absorbed energy.

Full working

The black plate warms up faster. A black, matt surface absorbs infrared radiation well and reflects very little. The absorbed energy is transferred to the thermal store of the plate, so its temperature rises quickly. The shiny plate reflects most of the infrared radiation, so it absorbs less energy and warms up slowly.

This answer is a model answer for you to compare with your own wording.

3. A ray of light hits a plane mirror. The angle between the incoming ray and the mirror is 30°. Work out the angle of reflection, and give a reason.

Hint

The angle of incidence is measured from the normal, not from the mirror. The normal is at right angles to the mirror.

Next step

Subtract 30° from 90° to find the angle of incidence. Then use the law of reflection.

Full working

The normal is at 90° to the mirror, so the angle of incidence is $90^\circ - 30^\circ = 60^\circ$ . The angle of reflection equals the angle of incidence, so the angle of reflection is 60°.

The reason is the law of reflection: the angle of reflection equals the angle of incidence.

4. A ray of light passes from air into a glass block. Describe what happens to the ray, and explain why.

Hint

Compare the speed of light in air with its speed in glass. Most of the light enters the glass, but a little does not.

Next step

Say which way the ray bends, then link the bend to the change of speed at the boundary.

Full working

Most of the light passes into the glass and bends towards the normal. A small part of the light is reflected at the surface. Light travels more slowly in glass than in air. The change of speed at the boundary is what makes the ray change direction.

5. A hospital sterilises sealed packs of equipment with gamma radiation, and images broken bones with X-rays. Explain why each kind of radiation is suitable for its job.

Hint

Think about how far each kind of radiation travels through matter, and what it does when it is absorbed.

Next step

For the gamma rays, say why they can reach inside a sealed pack. For the X-rays, say how bone and soft tissue interact differently with the radiation.

Full working

Gamma radiation is very penetrating, so it passes through the sealed pack and kills the microorganisms inside it. This sterilises the equipment without opening the pack.

X-rays pass through soft tissue but are absorbed by denser bone. The bone therefore casts a shadow on the detector, and the image shows the shape of the bone.

This answer is a model answer for you to compare with your own wording.