Question

1) the energy of a quantum light is 2.84 x 10^-19 J. What are the wavelength (nm) and the frequency (s^-1 ) of light?

2) What happens to the wavelength and the amplitude when $\lambda$ passes through one medium into another and out again? (i.e. air to solid to air)

Answer 1

Answer – 1) We are given energy of a quantum light is 2.84 x 10^-19 J. We need to calculate the he wavelength (nm) and the frequency (s^-1 ) of light.

We know formula for the energy and wavelength as follow –

E = h x C / λ

So, λ = h x C / E

Where, h = Planks constant, C = Speed of light, E = energy and λ = wavelength.

By plugging the value in the formula –

λ = 6.626 x 10^-34 J.s x 3.0 x 10⁸ m/s / 2.84 x 10^-19 J

= 7.00 x 10^-7 m

Now we need to convert the calculated wavelength m to m

We know

1 m = 1.0 x 10⁹ nm

So, 7.00 x 10^-7 m = ? nm

= 700 nm

So, wavelength of light is 700 nm

Now we need to calculate the frequency from the given energy and we know formula

E = h x v

Where , h = Planks constant, v = frequency

By rearranging the equation

v = E / h

       = 2.84 x 10^-19 J / 6.626 x 10^-34 J.s

      = 4.29 x 10¹⁴ s^-1.

So, frequency of light is 4.29 x 10¹⁴ s^-1.

2) We know when wave pass though one medium to another medium then there is wavelength and velocity both amplitude changed. When wave pass though one medium to another medium like air to solid mean rarer medium to denser medium then wave gets bend from its normal angle and there is lower index of refraction result increasing the wavelength. We also known as the wavelength increasing amplitude decrease, since wavelength is inversely proportional to energy and energy is directly proportional to amplitude. When again wave pass solid to air there is decrease the wavelength and increase the amplitude