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Chapter 6: Problem 29

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Short Answer

Expert verified
(a) The frequency of the photon is \(f = \frac{3.0 \times 10^8 m/s}{325 \times 10^{-9} m} = 9.23 \times 10^{14} Hz\). (b) The energy of a single photon is \(E = 6.63 \times 10^{-34} Js \cdot 9.23 \times 10^{14} Hz = 6.12 \times 10^{-19} J\). (c) The energy of one mole of photons is \(E_{mole} = 6.12 \times 10^{-19} J \cdot 6.022 \times 10^{23} entities/mol = 3.68 \times 10^5 J/mol\). (d) The number of photons in a 1.00 mJ burst of radiation is \(N_{photons} = \frac{1.00 \times 10^{-3} J}{6.12 \times 10^{-19} J} = 1.63 \times 10^{15} photons\). (e) The average energy of chemical bonds broken by the UV photons is \(E_{bond} = \frac{3.68 \times 10^{5} J/mol}{10^3} = 368\, kJ/mol\).

Step by step solution

01

1. Calculate the frequency of a photon with a wavelength of 325 nm

First, we need to determine the frequency of the photon with the given wavelength, 325 nm. We can use the speed of light equation for this, which is \(c = \lambda \cdot f \), where c is the speed of light (approximately \(3.0 \times 10^{8} m/s \)), \(\lambda\) is the wavelength (325 nm, but we must convert it to meters: \( 325 \times 10^{-9} m\)), and f is the frequency we want to find. By rearranging the equation, we get: \( f = \frac{c}{\lambda} = \frac{3.0 \times 10^8 m/s}{325 \times 10^{-9} m} \)
02

2. Calculate the energy of a single photon

Now we can use Planck's equation, \( E = h \cdot f \), to find the energy of a single photon. In this equation, E represents energy, h is Planck's constant (approximately \(6.63 \times 10^{-34} Js\)), and f is the frequency we found in step 1. Simply plug in the numbers: \( E = 6.63 \times 10^{-34} Js \cdot f \)
03

3. Calculate the energy of one mole of photons

Next, to find the energy of one mole of photons, we just need to multiply the energy of a single photon (found in step 2) by Avogadro's number, \( N_{A} \) (approximately \(6.022 \times 10^{23} entities/mol\)): \( E_{mole} = E \cdot N_{A} \)
04

4. Calculate the number of photons in a 1.00 mJ burst of radiation

In this step, we need to find out how many photons are in a 1.00 mJ burst of radiation. We can do this by dividing the total energy of the burst (1.00 mJ) by the energy of a single photon (found in step 2). First, we need to convert 1.00 mJ to Joules: \( 1.00 \, mJ = 1.00 \times 10^{-3} J \) Now, divide the total energy by the energy of a single photon: \( N_{photons} = \frac{1.00 \times 10^{-3} J}{E} \)
05

5. Estimate the average energy of chemical bonds in kJ/mol

Finally, to estimate the average energy of chemical bonds in the skin broken by the UV photons, we assume that each photon provides exactly the amount of energy needed to break one bond. We can use the energy of one mole of photons (found in step 3) and convert it from Joules to kJ/mol: \( E_{bond} = \frac{E_{mole}} { 10^3 } \, kJ/mol \) Now we can plug in the numbers for each step and solve the exercise.

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