# 0.7 Electromagnetic radiation  (Page 3/6)

 Page 3 / 6

## Wave nature of em radiation

1. List one source of electromagnetic waves. Hint: consider the spectrum diagram and look at the names we give to different wavelengths.
2. Explain how an EM wave propagates, with the aid of a diagram.
3. What is the speed of light? What symbol is used to refer to the speed of light? Does the speed of light change?
4. Do EM waves need a medium to travel through?

[link] lists the wavelength- and frequency ranges of the divisions of the electromagnetic spectrum.

 Category Range of Wavelengths (nm) Range of Frequencies (Hz) gamma rays $<$ 1 $>3×{10}^{19}$ X-rays 1-10 $3×{10}^{17}$ - $3×{10}^{19}$ ultraviolet light 10-400 $7,5×{10}^{14}$ - $3×{10}^{17}$ visible light 400-700 $4,3×{10}^{14}$ - $7,5×{10}^{14}$ infrared 700- ${10}^{5}$ $3×{10}^{12}$ - $4,3×{10}^{19}$ microwave ${10}^{5}-{10}^{8}$ $3×{10}^{9}$ - $3×{10}^{12}$ radio waves $>{10}^{8}$ $<3×{10}^{9}$

Examples of some uses of electromagnetic waves are shown in [link] .

 Category Uses gamma rays used to kill the bacteria in marshmallows and to sterilise medical equipment X-rays used to image bone structures ultraviolet light bees can see into the ultraviolet because flowers stand out more clearly at this frequency visible light used by humans to observe the world infrared night vision, heat sensors, laser metal cutting microwave microwave ovens, radar radio waves radio, television broadcasts

1. Arrange the following types of EM radiation in order of increasing frequency: infrared, X-rays, ultraviolet, visible, gamma.
2. Calculate the frequency of an EM wave with a wavelength of 400 nm.
3. Give an example of the use of each type of EM radiation, i.e. gamma rays, X-rays, ultraviolet light, visible light, infrared, microwave and radio and TV waves.

## The particle nature of electromagnetic radiation

When we talk of electromagnetic radiation as a particle, we refer to photons, which are packets of energy. The energy of the photon is related to the wavelength of electromagnetic radiation according to:

Planck's constant

Planck's constant is a physical constant named after Max Planck.

$h=6,626×{10}^{-34}$ J $·$ s

The energy of a photon can be calculated using the formula: $E=hf$ or $E=h\frac{c}{\lambda }$ . Where E is the energy of the photon in joules (J), h is planck's constant, c is the speed of light, f is the frequency in hertz (Hz) and $\lambda$ is the wavelength in metres (m).

Calculate the energy of a photon with a frequency of $3×{10}^{18}$  Hz

1. $\begin{array}{ccc}\hfill E& =& hf\hfill \\ & =& 6,6×{10}^{-34}×3×{10}^{18}\hfill \\ & =& 2×{10}^{-15}\phantom{\rule{0.166667em}{0ex}}\mathrm{J}\hfill \end{array}$

What is the energy of an ultraviolet photon with a wavelength of 200 nm?

1. We are required to calculate the energy associated with a photon of ultraviolet light with a wavelength of 200 nm.

We can use:

$E=h\frac{c}{\lambda }$
2. $\begin{array}{ccc}\hfill E& =& h\frac{c}{\lambda }\hfill \\ & =& \left(6,626×{10}^{-34}\right)\frac{3×{10}^{8}}{200×{10}^{-9}}\hfill \\ & =& 9,939×{10}^{-10}\phantom{\rule{0.166667em}{0ex}}\mathrm{J}\hfill \end{array}$

## Exercise - particle nature of em waves

1. How is the energy of a photon related to its frequency and wavelength?
2. Calculate the energy of a photon of EM radiation with a frequency of ${10}^{12}$  Hz.
3. Determine the energy of a photon of EM radiation with a wavelength of 600 nm.

if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
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John
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Grace
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John
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Grace
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acha se dhek ke bata sin theta ke value
Ajay
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Ajay
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Brittany
Prove that 4sin50-3tan 50=1
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f (x) = −3x + 5 and g (x) = x − 5 /−3
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@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
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24x^5
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10x
Axmed
24X^5
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how do you translate this in Algebraic Expressions
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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