Question

The wavelength of light that has a frequency of 1.20 1013s-1 is __________ m.
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Answer 1

Formula

The relationship between frequency and wavelength for an electromagnetic wave is given by

c=f λ———–(i)

Where,

c is the velocity of light

f is the frequency of light

λ is the wavelength of the light

Solution

To find the wavelength of the given light we will substitute the known values in equation (i) we get,

c= f λ

c= 3 X 108 ms-1

f= 1.20 × 1013 s-1

λ we need to find out

Therefore the equation becomes

3 X 108= 1.20 × 1013 x λ

λ= 3 X 108 / 1.20 × 1013

= 2.5 x 1o-5m

Hence, The wavelength of light that has a frequency of 1.20 × 1013 s-1 is 2.5 x 1o-5m

Answer 2

Answer:To find the wavelength of the given light we will substitute the known values in equation (i) we get,

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λ

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find out

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes3 X 108= 1.20 × 1013 x λ

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes3 X 108= 1.20 × 1013 x λλ= 3 X 108 / 1.20 × 1013

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes3 X 108= 1.20 × 1013 x λλ= 3 X 108 / 1.20 × 1013= 2.5 x 1o-5m

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes3 X 108= 1.20 × 1013 x λλ= 3 X 108 / 1.20 × 1013= 2.5 x 1o-5mHence, The wavelength of light that has a frequency of 1.20 × 1013 s-1 is 2.5 x 1o-5m

To find the wavelength of the given light we will substitute the known values in equation (i) we get,c= f λc= 3 X 108 ms-1f= 1.20 × 1013 s-1λ we need to find outTherefore the equation becomes3 X 108= 1.20 × 1013 x λλ= 3 X 108 / 1.20 × 1013= 2.5 x 1o-5mHence, The wavelength of light that has a frequency of 1.20 × 1013 s-1 is 2.5 x 1o-5m

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