Problem 6.29 One type of sunburn occurs on exposure to UV light of wavelength in

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Problem 6.29

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 300 nm .

Part A

What is the energy of a photon of this wavelength?

6.03•1019

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Part B

What is the energy of a mole of these photons?

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Part C

How many photons are in a 1.10 mJ burst of this radiation?

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Part D

These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 300 nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ/mol.

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Problem 6.29

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 300 nm .

Part A

What is the energy of a photon of this wavelength?

E =

6.03•1019

  J  

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Part B

What is the energy of a mole of these photons?

E =   J  

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Part C

How many photons are in a 1.10 mJ burst of this radiation?

N = photons

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Part D

These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 300 nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ/mol.

averagebondenergy =   kJ/mol  

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Explanation / Answer

A)

What is the energy of a photon of this wavelength?

WL = 300 nm

E = hc/WL

h = 6.626*10^-34 planck constant

c = 3*10^8 m/s speed of light

WL = wavelenght

E = (6.626*10^-34)(3*10^8)/(300*10^-9) = 6.63*10^-9 J per photon

B)

recall that

1 mol = 6.022*10^23 photons

therefore

E = e*NA = 6.63*10^-9 J per photon * 6.022*10^23 photons = 3.99*10^15 J per mol of photon

C)

6.63*10^-9 J --> 1 photon

1.19 mili J ---> 179487 photons

D)

This value must be the one in B, 1 mol of photon will provide the eneough energy to destroy 1 mol of bonds

3.99*10^15 J per mol of photon

but we require kJ so

3.99*10^12 kJ

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