IGCSE Physics 0625: Frequency of Ultraviolet Radiation Using v = fλ

A UV germicidal lamp in a water-purification plant emits ultraviolet radiation of wavelength $254\text{ nm}$ ( $2.54\times10^{-7}\text{ m}$ ) in air, where electromagnetic waves travel at $3.0\times10^{8}\text{ m/s}$ .

(a) Calculate the frequency of this ultraviolet radiation. Give your answer in standard form, to 3 significant figures. [2]

(b) State one everyday use of ultraviolet radiation, other than purifying water. [1]

(c) Visible red light has a longer wavelength than this ultraviolet radiation, at about $700\text{ nm}$ . State how the frequency of red light compares with the frequency of the ultraviolet radiation found in part (a), and explain your answer using the wave speed equation. [2]

Part (a): Calculating the frequency of the ultraviolet radiation

The wavelength is already given in metres: $\lambda = 2.54\times10^{-7}\text{ m}$ , and the speed is $v = 3.0\times10^{8}\text{ m/s}$ .

Rearrange $v = f\lambda$ to make frequency the subject:

$f = \frac{v}{\lambda}$

Substitute the values:

$f = \frac{3.0\times10^{8}}{2.54\times10^{-7}}$

$f = 1.181\ldots\times10^{15}\text{ Hz}$

Rounding to 3 significant figures:

$f = \boxed{1.18\times10^{15}\text{ Hz}}$

Part (b): A use of ultraviolet radiation

Ultraviolet radiation has several everyday uses besides water purification, including:

detecting forged banknotes, which contain fluorescent security features that only glow under UV light, or
sunbeds/tanning lamps, which use UV radiation to darken skin.

Any one correct use, other than water sterilisation, earns the mark.

Part (c): Comparing the frequency of red light with the ultraviolet radiation

Red light has a longer wavelength ( $700\text{ nm}$ ) than the ultraviolet radiation in part (a) ( $254\text{ nm}$ ). Since both waves travel through the same medium (air) at the same speed $v = 3.0\times10^{8}\text{ m/s}$ , the wave speed equation $v = f\lambda$ tells us that frequency and wavelength are inversely related when $v$ is fixed: a longer wavelength must go with a lower frequency, so that the product $f\lambda$ stays equal to $v$ .

Therefore, red light has a lower frequency than the ultraviolet radiation in part (a).

(As a check: $f_{\text{red}} = \dfrac{3.0\times10^{8}}{7.00\times10^{-7}} \approx 4.29\times10^{14}\text{ Hz}$ , which is indeed smaller than the ultraviolet frequency of $1.18\times10^{15}\text{ Hz}$ found in part (a).)

Final answers

(a) Frequency of the ultraviolet radiation $=$ $1.18\times10^{15}\text{ Hz}$
(b) Use of ultraviolet radiation $=$ e.g. detecting forged banknotes / sunbeds
(c) Frequency of red light $=$ lower than the ultraviolet radiation (longer wavelength, same speed, so $f=v/\lambda$ gives a smaller value)