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•Homework Set 4: Chapter 4 # 41 & 42 + Supplemental Problems

Light!

Frequency = n Wavelength = l Speed = c = lnEnergy E = hn

c = 299792458 m/s ≈ 3.00x108 m/s h = 6.626x10-34 J-s

Some simple examples•Determine the frequency of the red emission line of hydrogen whose wavelength is 656.3 nm

•Determine the energy of a sodium D1 photon whose wavelength is 589.592 nm

•What is the wavelength of an gamma-ray photon whose energy is 5.0 MeV? 1 eV = 1.602x10-19 Joules?

Example Solution

Given a wavelength of 656.3 nm, find the frequency.

814

9

2.9979 104.568 10

656.3 10

msc

c Hzm

Example Solution 2

34 8

199

19

19

6.626 10 2.99792 103.36914 10

589.592 10

3.370 10

, 2.103

ms

cc and E h

Jshcso E J

m

J

So E eV

A useful unit is the eV.

One ev is 1.60217657 10 J oules

Given a wavelength of 589.592 nm, find the

energy.

Example 3 SolutionGiven an energy of 5.0 MeV, find the wavelength.

First convert energy from MeV to Joules since h has units of J-s

34 8

1319

13

6.626 10 3.0 102.485 10

8.0109 10

2.5 10

msJ shc hc

E h mE J

m

6 19 195.0 10 1.60218 10 8.0109 10JeVE eV J

Now use energy formula to find wavelength

Light comes from electron transitions within the atom

For the hydrogen atom these transitions are named after the scientist who studied them. The Lyman lines are in the UV range, the Balmer are visible and the Paschen and Brackett are in the IR

The energy of an emitted photon is just the difference in

energy of two energy levels

22

116.13

if

initialfinal

nneV

hEEE

n is the quantum number for the initial or final energy level. Note that there is non = 0 energy level. The lowest energy level, n = 1, is referred to as the ground state.

For the hydrogen atom

ExamplesThe Balmer emission lines of hydrogen are transitions whose final energy level is the n = 2 level. Determine the wavelength, in nm, of the first four Balmer lines.

Example SolutionThe Balmer lines all have nf = 2 so the first four lines will have ni = 3, 4, 5 and 6. First do some algebra to get a formula for wavelength in terms of quantum numbers. Convert eV to J while we are at it.

192 2 2 2

181822 2

182

1 1 1 113.6 13.6 1.60218 10

1 11 1 2.17896 102.17896 104

1 12.17896 10

4

JeV

f i f i

if i

i

E eV eVn n n n

hc hc hc hcE

EJJ

nn n

so

hc

Jn

Example Solution 2Lump all the constants together since we have to do the same calculation several times

34 8

182 2

8

2

6.62607 10 2.99792 10

1 1 1 113.6 2.17896 10

4 4

9.116452 10

1 14

ms

i i

i

J shc

eV Jn n

m

n

Example Solution 3Now just plug in values for ni

8

2

87

1

2

87

2

2

87

3

2

8

4

2

9.116452 10

1 14

9.116452 106.5634 10 656

1 14 3

9.116452 104.8621 10 486

1 14 4

9.116452 104.341 10 434

1 14 5

9.116452 101 14 6

i

m

n

mm nm

mm nm

mm nm

m

74.1024 10 410m nm