Problems #11 - 20

Wavelength-Frequency-Energy Problems #1 - 10 | Go to Part Two of Light Equations |

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**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 10^{6} 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 10

^{6}eV = 1.670 x 10^{5}eV1.670 x 10

^{5}eV times 1.6022 x 10¯^{19}J/eV = 2.675674 x 10¯^{14}Juse E = hν:

2.675674 x 10¯

^{14}J = (6.6260755 x 10¯^{34}J s) (x)x = 4.038 x 10

^{19}s¯^{1}

**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 10^{8} m s¯^{1}

E = [ (6.626 x 10¯^{34}J s) (3.00 x 10^{8}m s¯^{1}) ] / 6.62 x 10¯^{7}mE = 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 Cl_{2} bonds. What is the wavelength of light (in nm) that would be required to break the bond of one Cl_{2} 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 = hcE = [ (6.626 x 10¯

^{34}J s) (3.00 x 10^{8}m s¯^{1}) ] / 5.32 x 10¯^{7}mE = 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 photon1.07 x 10

^{19}photons per second

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

Area = πr^{2}π(0.6)

^{2}/ π(3.1)^{2}0.03746

5) Determine photons through pinhole in one second:

(0.03746) (1.07 x 10^{19}) = 4.01 x 10^{17}

**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 10^{8}m/s)] / 4.30 x 10^{-19}Jλ = 4.6228 x 10

^{-7}mThis 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 = cto 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 10^{8}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 10^{10}cm s¯^{1})

Comment: I used Eλ = hc. Note also that I used 3.00 x 10^{10} 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 10^{23}mol¯^{1}) = 9.98 x 10^{8}J mol¯^{1}This value is usually reported in kJ per mole: 9.98 x 10

^{5}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 10^{6}eV(1.17 x 10

^{6}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 10^{23}mol¯^{1}) = 1.129 x 10^{11}J/mol = 1.129 x 10^{8}kJ/mol

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

**Example #19:** Light of wavelengths shorter than 275 nm can be used to photodissociate the hydrogen molecule into hydrogen atoms in the gas phase. A 60.0 mL glass cylinder contains H_{2}(g) at 45.0 mtorr and 25.0 °C . What minimum amount of light energy must be absorbed by the hydrogen in the tube to dissociate 38.0% of the molecules?

**Solution:**

1) Determine how many hydrogen molecules are present:

PV = nRT45.0 mtorr = 0.0450 torr

(0.0450 torr / 760.0 torr/atm) (0.0600 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 1.452788 x 10¯

^{7}mol(6.022 x 10

^{23}molecules/mol) (1.452788 x 10¯^{7}mol) = 8.74869 x 10^{16}molecules

2) Determine 38.0%:

(0.380) (8.74869 x 10^{16}molecules) = 3.3245 x 10^{16}molecules

3) Determine energy of one photon with a wavelength of 275 nm:

Eλ = hc(E) (2.75 x 10¯

^{7}m) = (6.626 x 10¯^{34}J s) (3.00 x 10^{8}m/s)E = 7.22836 x 10¯

^{19}J (per molecule of H_{2})

4) Determine minimum energy required for the 38.0%:

(7.22836 x 10¯^{19}J) (3.3245 x 10^{16}) = 0.0240 J

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 |