Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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

ID: 909921 • Letter: P

Question

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?

6.6•109

SubmitMy AnswersGive Up

Incorrect; Try Again; 2 attempts remaining

Part B

What is the energy of a mole of these photons?

4•1015

SubmitMy AnswersGive Up

Incorrect; Try Again; 4 attempts remaining

Part C

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

179487

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

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.

3.99•1012

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

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 =

6.6•109

  J  

SubmitMy AnswersGive Up

Incorrect; Try Again; 2 attempts remaining

Part B

What is the energy of a mole of these photons?

E =

4•1015

  J  

SubmitMy AnswersGive Up

Incorrect; Try Again; 4 attempts remaining

Part C

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

N =

179487

photons

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

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 =

3.99•1012

  kJ/mol  

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

WL = 300 nm

A)

Energy

E = hc/WL

h = 6.626*10^-34 planck constant

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

E = (6.626*10^-34)(3*10^8) / (300*10^-9) = 6.626e*10^-19

B)

Energy of 1 mol of photon

E =  6.626e*10^-19 * 6.022*10^23 = 1084642.61751 J

C)

if E = 1.1 *10^-3 J

find numebr of photons

6.626e*10^-19 J/e-

then

#photon = (1.1 *10^-3 J)/(6.626e*10^-19 J/p) = 1.66*10-15 photons