Electromagnetic Waves

Solution

1. 1

   Recall the AQA mnemonic R-M-I-V-U-X-G in the longest-to-shortest direction.

2. 2

   Write the seven groups in that order.

   $\text{radio} \to \text{microwave} \to \text{infrared} \to \text{visible} \to \text{ultraviolet} \to \text{X-ray} \to \text{gamma}$

Solution

1. 1

   Use the wave equation.

   $c = f \lambda$

2. 2

   Rearrange for frequency.

   $f = \dfrac{c}{\lambda}$

3. 3

   Substitute and evaluate.

   $f = \dfrac{3.0\times10^8}{3.0} = 1.0 \times 10^8\,\text{Hz}$

4. 4

   Convert to MHz for context.

   $1.0\times10^8\,\text{Hz} = 100\,\text{MHz}$

Solution

1. 1

   State the visible range.

   $\text{eye detects } \sim 400\text{ nm to } \sim 700\text{ nm}$

2. 2

   Compare ultraviolet's wavelength with this range.

   $\lambda_{\text{UV}} < 400\,\text{nm}$

3. 3

   Conclude using the retina's response.

Solution

1. 1

   Recognise the direction of the speed change.

   $v_{\text{glass}} < v_{\text{air}}$

2. 2

   Apply the refraction rule for entering a denser medium.

   $\text{ray bends towards the normal} \Rightarrow r < i$

3. 3

   State the conclusion explicitly.

   $r < 50°$

Solution

1. 1

   Describe the second refraction event.

   $\text{glass} \to \text{air}: \; v \text{ increases}$

2. 2

   State the direction of bending using the normal.

   $\text{ray bends away from the normal}$

3. 3

   Conclude by comparing the two faces (they are parallel).

   $\text{angles cancel out} \Rightarrow \text{emerging ray parallel to incident ray}$

Solution

1. 1

   Use $v = f\lambda$ for the air.

   $\lambda_{\text{air}} = \dfrac{v}{f} = \dfrac{3.0\times10^8}{5.0\times10^{14}}$

2. 2

   Evaluate.

   $\lambda_{\text{air}} = 6.0\times10^{-7}\,\text{m} = 600\,\text{nm}$

3. 3

   Repeat for the glass (same frequency).

   $\lambda_{\text{glass}} = \dfrac{2.0\times10^8}{5.0\times10^{14}} = 4.0\times10^{-7}\,\text{m} = 400\,\text{nm}$

4. 4

   Comment.

   $f \text{ unchanged}; \; v, \lambda \text{ decrease in glass}$

Solution

1. 1

   The oscillating electrons in the aerial radiate an EM wave with the same frequency as the AC.

   $f = 98.7\,\text{MHz} = 98.7\times10^6\,\text{Hz}$

2. 2

   Use the wave equation v = fλ rearranged for λ.

   $\lambda = \dfrac{v}{f} = \dfrac{3.0\times10^8}{98.7\times10^6}$

3. 3

   Evaluate.

   $\lambda \approx 3.04\,\text{m}$

Solution

1. 1

   The radio wave is an oscillating electric and magnetic field that arrives at the aerial.

2. 2

   The oscillating electric field exerts a force on the free electrons in the metal aerial.

3. 3

   The electrons are forced to oscillate up and down the aerial at the same frequency as the wave, producing an alternating current that the receiver electronics amplify.

Solution

1. 1

   (a) Visible light: an electron in a hot carbon atom falls from a higher to a lower energy level and emits a photon of visible light.

2. 2

   (b) X-rays: high-speed electrons hit a metal anode and decelerate rapidly; the deceleration produces X-ray photons.

3. 3

   (c) Gamma rays: changes inside the cobalt-60 nucleus after beta decay leave the nucleus in an excited state; rearrangement releases a gamma photon.

Solution

1. 1

   Divide the annual background by the dose per X-ray.

   $n = \dfrac{2.7\,\text{mSv}}{0.014\,\text{mSv}}$

2. 2

   Evaluate.

   $n \approx 193$

Solution

1. 1

   UV photons have enough energy to ionise atoms / molecules in skin cells (1 mark).

2. 2

   Ionisation damages DNA, which can cause uncontrolled cell division (a tumour / cancer) (1 mark).

Solution

1. 1

   Calculate the wavelength: λ = v/f = (3×10⁸)/(1.98×10⁵) ≈ 1.5 km.

2. 2

   Long-wavelength waves diffract strongly around obstacles such as hills.

3. 3

   Short-wavelength TV signals diffract very little and need line of sight; long-wave radio bends around hills to reach hidden glens.

Solution

1. 1

   The microwave frequency (about 2.45 GHz) is chosen to be strongly absorbed by water molecules.

2. 2

   Water molecules in food gain kinetic energy as they absorb the microwaves, so the food heats up.

3. 3

   The plastic dish contains no free water, so the microwaves pass through it without being absorbed and the dish stays cool.

Solution

1. 1

   X-rays are directed through the patient onto a digital detector behind the body.

2. 2

   Soft tissue (lung, skin) absorbs very little — X-rays pass through, exposing the detector behind those areas (dark on the image).

3. 3

   Dense bone absorbs much more X-ray energy. The detector behind the bone is shielded → those areas appear light, giving the rib shadow.

Solution

1. 1

   Visible / IR wavelengths are < 1 µm, comparable to or smaller than the fibre core, so they undergo total internal reflection along the fibre.

2. 2

   Glass is highly transparent at near-IR wavelengths — very low energy loss per kilometre, allowing long-distance data transmission.

3. 3

   Microwaves have wavelengths of cm or more, far larger than the fibre core. They would not undergo total internal reflection and would not propagate down the fibre.

Solution

1. 1

   Gamma rays are very penetrating — they pass easily through plastic packaging to reach the contents.

2. 2

   Gamma rays are ionising — they damage the DNA of any bacteria inside, killing them.

3. 3

   Bonus: this happens at room temperature, so heat-sensitive items inside the packet are not affected, and the scalpels remain sealed.