The protective action of ozone in the atmosphere comes through ozone's absorption of UV radiation in the 230 to 290 nm wavelength range. What is the energy, in kilojoules per mole, associated with radiation in the wavelength range?
520-413kj/mol
Oh, so you want to get technical, huh? Well, let me juggle some numbers for you. The energy associated with radiation in the 230 to 290 nm wavelength range is approximately 146 to 190 kilojoules per mole. That's enough energy to power a spontaneous burst of laughs! Now that's some electrifying comedy, don't you think?
To find the energy associated with radiation in the given wavelength range, we can use the formula:
E = hc/λ
where:
E is the energy of the radiation,
h is Planck's constant (6.626 × 10^-34 J*s),
c is the speed of light (2.998 × 10^8 m/s), and
λ is the wavelength of the radiation.
First, we need to convert the wavelength range from nm to meters. Since 1 nm = 1 × 10^-9 meters, we have:
λ₁ = 230 nm = 230 × 10^-9 m
λ₂ = 290 nm = 290 × 10^-9 m
Now we can find the energy using the formula:
E₁ = hc/λ₁
E₂ = hc/λ₂
Let's calculate the energy for λ₁:
E₁ = (6.626 × 10^-34 J*s * 2.998 × 10^8 m/s) / (230 × 10^-9 m)
Simplifying:
E₁ = 8.613 × 10^-19 J
To convert the energy from joules to kilojoules, we divide by 1000:
E₁ = 8.613 × 10^-22 kilojoules
Similarly, let's calculate the energy for λ₂:
E₂ = (6.626 × 10^-34 J*s * 2.998 × 10^8 m/s) / (290 × 10^-9 m)
Simplifying:
E₂ = 6.877 × 10^-19 J
Converting to kilojoules:
E₂ = 6.877 × 10^-22 kilojoules
Therefore, the energy associated with radiation in the 230 to 290 nm wavelength range is approximately 8.613 × 10^-22 to 6.877 × 10^-22 kilojoules per mole.
413.11 kJ/mol
E = hc/wavelength
Remember h is Planck's constant
c is speed of light in m/s
wavelength must be in m
Solve for E in J and this is for 1 photon. Multiply by 6.02E23 to obtain J/mol and convert to kJ.
Do this for 230 nm and 290 nm and I would quote the range in energy.