The Electromagnetic Spectrum: Question 5

Syllabus 3.3

Structured Extended 7 marks

A mobile-phone signal has frequency 1800 MHz1800\text{ MHz}. An FM radio broadcast has wavelength 3.2 m3.2\text{ m}. Both signals travel through air at 3.0×108 m/s3.0\times10^{8}\text{ m/s}.

(a) Calculate the wavelength of the mobile-phone signal. Give your answer in metres, to 2 significant figures. [2]

(b) Calculate the frequency of the FM radio signal. Give your answer in standard form, to 2 significant figures. [2]

(c) State which of the two signals, the mobile-phone signal or the FM radio signal, has the higher frequency and the shorter wavelength. [1]

(d) Both signals lie towards the low-frequency end of the electromagnetic spectrum, where the radiation is non-ionising. Explain why exposure to radio waves and microwaves at typical everyday intensities is not considered as harmful as exposure to ultraviolet or X-ray radiation. [2]

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Worked solution

Part (a): Wavelength of the mobile-phone signal

Convert the frequency to hertz:

f=1800 MHz=1800×106 Hz=1.8×109 Hzf = 1800\text{ MHz} = 1800\times10^{6}\text{ Hz} = 1.8\times10^{9}\text{ Hz}

Rearrange v=fλv = f\lambda to make wavelength the subject, and substitute:

λ=vf=3.0×1081.8×109\lambda = \frac{v}{f} = \frac{3.0\times10^{8}}{1.8\times10^{9}}

λ=0.16 m0.17 m\lambda = 0.1\overline{6}\text{ m} \approx \boxed{0.17\text{ m}}

Part (b): Frequency of the FM radio signal

Rearrange v=fλv = f\lambda to make frequency the subject, and substitute λ=3.2 m\lambda = 3.2\text{ m}:

f=vλ=3.0×1083.2f = \frac{v}{\lambda} = \frac{3.0\times10^{8}}{3.2}

f=9.375×107 Hz9.4×107 Hzf = 9.375\times10^{7}\text{ Hz} \approx \boxed{9.4\times10^{7}\text{ Hz}}

Part (c): Comparing the two signals

The mobile-phone signal has frequency 1.8×109 Hz1.8\times10^{9}\text{ Hz}, while the FM radio signal has frequency 9.4×107 Hz9.4\times10^{7}\text{ Hz}. Since 1.8×109>9.4×1071.8\times10^{9} > 9.4\times10^{7}, the mobile-phone signal has the higher frequency.

Because v=fλv = f\lambda is constant for both (they travel at the same speed), a higher frequency corresponds to a shorter wavelength. This is confirmed by the wavelengths found above: 0.17 m0.17\text{ m} (mobile phone) is much shorter than 3.2 m3.2\text{ m} (FM radio).

So the mobile-phone signal has both the higher frequency and the shorter wavelength.

Part (d): Why low-frequency electromagnetic waves are less hazardous

Radio waves and microwaves sit at the low-frequency, low-energy end of the electromagnetic spectrum. The energy carried by an electromagnetic wave increases as its frequency increases, so radio waves and microwaves carry far less energy per wave than the high-frequency waves at the other end of the spectrum, such as ultraviolet, X-rays and gamma rays.

Ultraviolet, X-rays and gamma rays carry enough energy to be ionising: they can knock electrons out of atoms and break chemical bonds, which is what allows them to damage or mutate living cells (for example, causing skin damage or cancer). Radio waves and microwaves do not carry enough energy to ionise atoms or break these bonds, so they are non-ionising and do not cause this kind of cellular damage at typical exposure levels.

(This does not mean radio waves and microwaves are hazard-free in every situation (microwaves can heat body tissue if exposed at very high intensity, which is a thermal effect rather than ionising damage) but this thermal risk is much less severe than the cell-damaging effects of ionising radiation.)

Final answers

  • (a) Mobile-phone wavelength == 0.17 m0.17\text{ m}
  • (b) FM radio frequency == 9.4×107 Hz9.4\times10^{7}\text{ Hz}
  • (c) Higher frequency, shorter wavelength == the mobile-phone signal
  • (d) Radio waves and microwaves are non-ionising (much lower energy per wave than UV/X-rays), so they cannot ionise atoms or break bonds in cells the way ultraviolet and X-rays can