Respuesta :
Answer:
b) twice the energy of each photon of the red light.
Explanation:
[tex]\lambda[/tex] = Wavelength
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
Energy of a photon is given by
[tex]E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}[/tex]
Let [tex]\lambda_1[/tex] = 700 nm
[tex]\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}[/tex]
For red light
[tex]E_1=\dfrac{hc}{\lambda_1}[/tex]
For UV light
[tex]E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}[/tex]
Dividing the equations
[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1[/tex]
Hence, the answer is b) twice the energy of each photon of the red light.
Answer:
b) twice the energy of each photon of the red light.
Explanation:
Given:
wavelength of red light, [tex]\lambda_r=7\times 10^{-7}\ m[/tex]
wavelength of ultraviolet light, [tex]\lambda_{uv}=3.5\times 10^{-7}\ m[/tex]
We know that the energy of a photon is given as:
[tex]E=h.\nu[/tex] ..............(1)
where:
h = plank's constant [tex]=6.626\times 10^{-34}\ J.s[/tex]
[tex]\nu=[/tex] frequency of the wave
we have the relation:
[tex]\nu=\frac{c}{\lambda}[/tex] ...................(2)
From (1) and (2) we have:
[tex]E=\frac{h.c}{\lambda}[/tex]
Energy of photons for ultraviolet light:
[tex]E_{uv}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.5\times 10^{-7}}[/tex]
[tex]E_{uv}=5.6794\times 10^{-19}\ J[/tex]
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.
