Calculations between wavelength, frequency and energy
Problems #11 - 20

Wavelength-Frequency-Energy Problems #1 - 10      Go to Part Two of Light Equations
Return to Part One of Light Equations      Return to Electrons in Atoms menu

Problem #11: The radioactive isotope Thallium-201 is used in medical diagnosis and treatment. A gamma ray emitted by an atom of Thallium-201 has an energy of 0.1670 million electron-volts. (1 MeV is 1 x 106 eV and 1 eV = 1.6022 x 10¯19 J). What is the frequency in Hz of this gamma ray?

Solution:

The first part converts the MeV value into Joules:

0.1670 MeV = 0.1670 x 106 eV = 1.670 x 105 eV

1.670 x 105 eV times 1.6022 x 10¯19 J/eV = 2.675674 x 10¯14 J

use E = hν:

2.675674 x 10¯14 J = (6.6260755 x 10¯34 J s) (x)

x = 4.038 x 10191


Problem #12: What is the energy of light with a wavelength of 662 nm?

Solution:

Convert wavelength to meters:

662 nm = 662 x 10¯9 m = 6.62 x 10¯7 m

Use Eλ = hc (with the speed of light as 3.00 x 108 m s¯1

E = [ (6.626 x 10¯34 J s) (3.00 x 108 m s¯1) ] / 6.62 x 10¯7 m

E = 3.00 x 10¯19 J

This is the amount of energy per photon.

Often, this type of question will ask for the energy in kJ/mol. In that case, multiply by Avogadro's Number, the divide by 1000 to arrive at the answer.


Problem #13: It requires 325 kJ to break one mole of Cl2 bonds. What is the wavelength of light (in nm) that would be required to break the bond of one Cl2 molecule.


Problem #14: An argon ion laser puts out 4.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 6.2 mm. If the laser is pointed toward a pinhole with a diameter of 1.2 mm, how many photons will travel through the pinhole per second? Assume that the light intensity is equally distributed throughout the entire cross-sectional area of the beam. (1 W = 1 J/s)

Solution:

1) Convert nm to m:

532 nm = 532 x 10¯9 m = 5.32 x 10¯7 m

2) Determine energy of one photon at this wavelength:

Eλ = hc

E = [ (6.626 x 10¯34 J s) (3.00 x 108 m s¯1) ] / 5.32 x 10¯7 m

E = 3.736466 x 10¯19 J

3) Determine power output in terms of photons:

4 J s¯1 / 3.736466 x 10¯19 J per photon

1.07 x 1019 photons per second

4) Determine decimal percent pinhole is of total area of beam:

Area = πr2

π(0.6)2 / π(3.1)2

0.03746

5) Determine photons through pinhole in one second:

(0.03746) (1.07 x 1019) = 4.01 x 1017

Problem #15: When a cesium salt solution is ionized in a Bunsen or Meeker burner, photons of energy 4.30 x 10-19 J are emitted. What color is the cesium flame?

Solution:

1) We combine E = hν and λν = c to arrive at:

λ = hc / E

2) We solve for the wavelength:

λ = [(6.626 x 10-34 J s) (3.00 x 108 m/s)] / 4.30 x 10-19 J

λ = 4.6228 x 10-7 m

This is 462.3 nm (You may do the conversion to check it.)

3) To determine the color, we look at a chart of wavelengths and colors. Here is one:

Note that this chart is divided up into seven colors (the well-known Roy G. Biv). There are many who disagree that indigo deserves to be a color in the spectrum, so choosing blue as the answer seems the best way to go.


Problem #16: The energy of a particular color of green light is 3.82 x 10-22 kJ. What is its wavelength in nanometers?

Solution:

We will assume the energy given is for one photon.

1) Convert kJ to J:

3.82 x 10-22 kJ times (1000 J / 1 kJ) = 3.82 x 10-19 J

2) What you now do is combine these two equations:

E = hν and λν = c

to eliminate the frequency. You wind up with this:

Eλ = hc

3) Calculate the wavelength in meters:

(3.82 x 10-19 J) (λ) = (6.626 x 10-34 J s) (3.00 x 108 m/s)

λ = 5.20 x 10-7 m

4) You may do the math set up for converting from meters to nanometers. The answer will be 520. nm.

By the way, Eλ = hc can also be written as E = hc/λ. It's just a stylistic thing to write it the first way.


Example #17: A particular x-ray has a wavelength of 1.2 Å. Calculate the energy of one mole of photons with this wavelength.

Solution:

1.2 Å x (10¯8 cm / 1 Å) = 1.2 x 10¯8 cm

(x) (1.2 x 10¯8 cm) = (6.626 x 10¯34 J s) (3.00 x 1010 cm s¯1)

Comment: I used Eλ = hc. Note also that I used 3.00 x 1010 cm s¯1 for the speed of light. I did this because the 1.2 Ångstrom value for the wavelength converts very easily into cm. There was no need to take the wavelength to meters.

x = 1.66 x 10¯15 J

This is the energy for one photon. To get energy per mole, multiply the above value by Avogadro's Number:

(1.66 x 10¯15 J) (6.022 x 1023 mol¯1) = 9.98 x 108 J mol¯1

This value is usually reported in kJ per mole: 9.98 x 105 kJ mol¯1


Example #18: When it decays by beta decay, cobalt-60 also produces two gamma rays with energies of 1.17 MeV and 1.33 MeV. For the 1.17 MeV photon, determine (a) the energy in Joules of one photon as well as (b) the energy produced in units of kJ per mole.

Solution:

To solve this problem, you must know the relationship between electron-volts (eV) and Joules. This value can be looked up. I will use 1 eV = 1.6022 x 10¯19 J for this problem.

(a) convert MeV to Joules:

1.17 MeV = 1.17 x 106 eV

(1.17 x 106 eV) x (1.6022 x 10¯19 J/eV) = 1.87 x 10¯13 J (to three sig. figs.)

(b) convert J of one photon to kJ/mol of photons:

(1.874574 x 10¯13 J) x (6.022 x 1023 mol¯1) = 1.129 x 1011 J/mol = 1.129 x 108 kJ/mol

Note the use of the unrounded off value for the energy of one photon.


Wavelength-Frequency-Energy Problems #1 - 10      Go to Part Two of Light Equations
Return to Part One of Light Equations      Return to Electrons in Atoms menu