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Solved: Excessive exposure to sunlight increases the risk of skin cancer because some of
Chapter 9, Problem 113P(choose chapter or problem)
Excessive exposure to sunlight increases the risk of skin cancer because some of the photons have enough energy to break chemical bonds in biological molecules. These bonds require approximately 250–800 kJ/mol of energy to break. The energy of a single photon is given by
𝘌 = 𝘩𝘤/l, where E is the energy of the photon in J, 𝘩 is Planck’s constant (\(6.626 \times 10^{-34} \mathrm{~J}\)s), and c is the speed of light (\(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\)). Determine which kinds of light contain enough energy to break chemical bonds in biological molecules by calculating the total energy in 1 mol of photons for light of each wavelength.
(a) infrared light (\(1500 nm\))
(b) visible light (\(500 nm\))
(c) ultraviolet light (\(150 nm\))
Equation Transcription:
6.626 10-34J
3.00 × 108 m/s
Text Transcription:
6.626 10^-34J
3.00 × 10^8 m/s
1500 nm
500 nm
150 nm
Questions & Answers
QUESTION:
Excessive exposure to sunlight increases the risk of skin cancer because some of the photons have enough energy to break chemical bonds in biological molecules. These bonds require approximately 250–800 kJ/mol of energy to break. The energy of a single photon is given by
𝘌 = 𝘩𝘤/l, where E is the energy of the photon in J, 𝘩 is Planck’s constant (\(6.626 \times 10^{-34} \mathrm{~J}\)s), and c is the speed of light (\(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\)). Determine which kinds of light contain enough energy to break chemical bonds in biological molecules by calculating the total energy in 1 mol of photons for light of each wavelength.
(a) infrared light (\(1500 nm\))
(b) visible light (\(500 nm\))
(c) ultraviolet light (\(150 nm\))
Equation Transcription:
6.626 10-34J
3.00 × 108 m/s
Text Transcription:
6.626 10^-34J
3.00 × 10^8 m/s
1500 nm
500 nm
150 nm
ANSWER:
Solution 113P
Here, we are going to calculate the total energy in 1 mol of photons for light of each wavelength.
Step 1:
- The energy of a single photon = hc/ƛ
Therefore, energy of 1 mol of photons = NA hc/ƛ
Given, wavelength(ƛ) = 1500 nm(infrared light) = 1500 x 10-9 m = 1.5 x 10-6 m [1 nm = 10-9 m]
h = 6.626 x 10-34 J.s
C = 3 x 108 m/s
NA = 6.02 x 1023
Therefore, energy of 1 mol of photons = [(6.02 x 1023)(6.626 x 10-34 J.s)(3 x 108 m/s)] / (1.5 x 10-6 m)
= 79.78 x 103 J
= 79.8 kJ
b) Given, wavelength(ƛ) = 500 nm(visible light) = 500 x 10-9 m = 5 x 10-7 m [1 nm = 10-9 m]
h = 6.626 x 10-34 J.s
C = 3 x 108 m/s
NA = 6.02 x 1023
Therefore, energy of 1 mol of photons = [(6.02 x 1023)(6.626 x 10-34 J.s)(3 x 108 m/s)] / (5 x 10-7 m)
= 23.93 x 104 J
= 239.3 kJ
c) Given, wavelength(ƛ) = 150 nm(ultraviolet light) = 150 x 10-9 m = 1.50 x 10-7 m [1 nm = 10-9 m]
h = 6.626 x 10-34 J.s
C = 3 x 108 m/s
NA = 6.02 x 1023
Therefore, energy of 1 mol of photons = [(6.02 x 1023)(6.626 x 10-34 J.s)(3 x 108 m/s)] / (1.50 x 10-7 m)
= 79.78 x 104 J
= 797.8 kJ