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Paul [167]
4 years ago
9

Electromagnetic radiation can be specified by its wavelength ( ), its frequency ( ) or its period ( ). The period is the time it

takes one complete wavelength to pass a point in space. Based on this information, what is the mathematical relationship between and ?
Physics
1 answer:
kirill [66]4 years ago
3 0

Answer:

The relation between f,T,λ

f=\frac{1}{T}

λ =\frac{v}{f}

λ =\frac{v}{T}

Explanation:

Let us consider,

                         wavelength as λ

                             frequency as  f

                             time period as T

The <em>time period</em> is defined as time taken by the wave to complete one oscillation or number or seconds to complete one full oscillation.

The <em>frequency </em>is defined as number of oscillation done per second.

The <em>wavelength</em> is defined as distance between two successive crests or trough. in a wave.

<em>The frequency and time period are reciprocals. </em>

<em>The wavelength and frequency are reciprocals ie larger the wavelength shorter is the frequency</em>

The relation between frequency and time period is f =\frac{1}{T}.

The relation between wavelength and frequency is λ =\frac{v}{f}

                     where v is velocity of light.

The relation between wavelength and time period is λ =\frac{v}{T}

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What is the wavelength, in nm, of a photon with energy
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Answer:

(a)  λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

E = h\frac{c}{\lambda}\\\\ \lambda = \frac{hc}{E}

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m

λ = 41.36 x 10⁻⁹ m

λ = 41.36 nm →ultraviolet

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3 years ago
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