How much greater is the energy of a photon of ultraviolet radiation (λ = 3.00 × 10-7 m) than the energy of an average photon of sunlight (λ = 5.60 × 10-7 m)?
1.87
Well, the energy of a photon is proportional to its frequency, which is inversely proportional to its wavelength. So, since ultraviolet radiation has a shorter wavelength than sunlight, it's like comparing a short joke to a long joke. The punchline (energy) of the short joke is greater because it gets to the point quicker, just like the photon of ultraviolet radiation has more energy than the average photon of sunlight. So, the energy of the ultraviolet photon is greater by "wavelengths"!
To calculate the energy of a photon, we can use the equation:
E = hc/λ
Where:
E is the energy of the photon,
h is Planck's constant (6.626 × 10^-34 J.s),
c is the speed of light (3.00 × 10^8 m/s),
λ is the wavelength of the photon.
Let's calculate the energy of the ultraviolet photon:
E_uv = (6.626 × 10^-34 J.s * 3.00 × 10^8 m/s) / (3.00 × 10^-7 m)
E_uv = 6.626 × 10^-19 J
Now, let's calculate the energy of the average sunlight photon:
E_sun = (6.626 × 10^-34 J.s * 3.00 × 10^8 m/s) / (5.60 × 10^-7 m)
E_sun = 3.54 × 10^-19 J
Now, let's find the difference between the two energies:
Difference = E_uv - E_sun
Difference = (6.626 × 10^-19 J) - (3.54 × 10^-19 J)
Difference = 3.086 × 10^-19 J
Therefore, the energy of a photon of ultraviolet radiation is 3.086 × 10^-19 Joules greater than the energy of an average photon of sunlight.
To calculate the energy of a photon, we can use the equation:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength of the photon.
Let's calculate the energy of a photon of ultraviolet radiation:
E_uv = hc/λ_uv
E_uv = (6.626 × 10^-34 J·s) × (3.00 × 10^8 m/s) / (3.00 × 10^-7 m)
Simplifying further:
E_uv = (6.626 × 3.00) × (10^-34 × 10^8) / 3.00 × 10^-7
E_uv = 19.878 × 10^-34 / 3.00 × 10^-7
E_uv = 6.626 × 19.878 / 3
E_uv ≈ 34.60 × 10^-34 J
Now, let's calculate the energy of an average photon of sunlight:
E_sun = hc / λ_sun
E_sun = (6.626 × 10^-34 J·s) × (3.00 × 10^8 m/s) / (5.60 × 10^-7 m)
Simplifying further:
E_sun = (6.626 × 3.00) × (10^-34 × 10^8) / 5.60 × 10^-7
E_sun = 19.878 × 10^-34 / 5.60 × 10^-7
E_sun = 6.626 × 19.878 / 5.60
E_sun ≈ 23.42 × 10^-34 J
To find how much greater the energy of a photon of ultraviolet radiation is compared to an average photon of sunlight, we can subtract the energy of the sunlight photon from the energy of the ultraviolet photon:
Difference = E_uv - E_sun
Difference = 34.60 × 10^-34 J - 23.42 × 10^-34 J
Difference ≈ 11.18 × 10^-34 J
Therefore, the energy of a photon of ultraviolet radiation is approximately 11.18 × 10^-34 J greater than the energy of an average photon of sunlight.
Note: In this calculation, we assumed that "average sunlight" refers to the visible light spectrum, which is roughly between 400-700 nm (4.00 × 10^-7 m to 7.00 × 10^-7 m).