Question

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at a wavelength of 350 nm. Each photon of the ultraviolet light has:________.

a) half the energy of each photon of the red light.
b) twice the energy of each photon of the red light.
c) four times the energy of each photon of the red light.
d) either more or less energy than each photon of the red light, depending on the intensities of the two light sources.

Answer 1

b) twice the energy of each photon of the red light.

Step-by-step explanation:

Given:

wavelength of red light,
`\lambda_r=7* 10^(-7)\ m`

wavelength of ultraviolet light,
`\lambda_(uv)=3.5* 10^(-7)\ m`

We know that the energy of a photon is given as:

`E=h.\\u` ..............(1)

where:

h = plank's constant
`=6.626* 10^(-34)\ J.s`

`\\u=` frequency of the wave

we have the relation:

`\\u=(c)/(\lambda)` ...................(2)

From (1) and (2) we have:

`E=(h.c)/(\lambda)`

Energy of photons for ultraviolet light:

`E_(uv)=(6.626* 10^(-34)* 3* 10^8)/(3.5* 10^(-7))`

`E_(uv)=5.6794* 10^(-19)\ J`

Since the energy of photons is inversely proportional to the wavelength of the light hence the photon-energy of ultraviolet light is double of the photon-energy of the red light as the wavelength of the red light is twice to that of ultraviolet light.

Answer 2

b) twice the energy of each photon of the red light.

Step-by-step explanation:

`\lambda` = Wavelength

h = Planck's constant =
`6.626* 10^(-34)\ m^2kg/s`

c = Speed of light =
`3* 10^8\ m/s`

Energy of a photon is given by

`E=h\\u\\\Rightarrow E=h(c)/(\lambda)`

Let
`\lambda_1` = 700 nm

`\lambda_2=350\\\Rightarrow \lambda_2=(\lambda_1)/(2)`

For red light

`E_1=(hc)/(\lambda_1)`

For UV light

`E_2=(hc)/((\lambda_1)/(2))`

Dividing the equations

`(E_1)/(E_2)=((hc)/(\lambda_1))/((hc)/((\lambda_1)/(2)))\\\Rightarrow (E_1)/(E_2)=(1)/(2)\\\Rightarrow E_2=2E_1`

Hence, the answer is b) twice the energy of each photon of the red light.