Problems #1 - 10

Wavelength-Frequency-Energy Problems #11 - 20 | Go to Part Two of Light Equations |

Return to Part One of Light Equations | Return to Electrons in Atoms menu |

**Problem #1:** A certain source emits radiation of wavelength 500.0 nm. What is the energy, in kJ, of one mole of photons of this radiation?

**Solution:**

1) Convert nm to m:

500.0 nm = 500.0 x 10¯^{9}m = 5.000 x 10¯^{7}m

2) Determine the frequency:

λν = c(5.000 x 10¯

^{7}m) (x) = 3.00 x 10^{8}m/sx = 6.00 x 10

^{14}s¯^{1}

3) Determine the energy:

E = hν:x = (6.626 x 10¯

^{34}J s) (6.00 x 10^{14}s¯^{1})x = 3.9756 x 10¯

^{19}JImportant point: this is the energy for one photon.

4) Determine energy for one mole of photons:

(3.9756 x 10¯^{19}J) (6.022 x 10^{23}mol¯^{1})239.4 kJ/mol

Comment: if you wished to do a direct calculation, you could use this equation:

E = hc / λ

Just make sure that the units for c and λ are the same.

**Problem #2:** If it takes 3.36 x 10^{-19} J of energy to eject an electron from the surface of a certain metal, calculate the longest possible wavelength, in nanometers, of light that can ionize the metal.

**Solution:**

1) Determine the frequency:

E = hν3.36 x 10

^{-19}J = (6.626 x 10¯^{34}J s) (x)x = 5.071 x 10

^{14}s¯^{1}

2) Determine the wavelength:

λν = c(x) (5.071 x 10

^{14}s¯^{1}) = 3.00 x 10^{8}m/sx = 5.916 x 10¯

^{7}m592 nm

**Problem #3:** Determine the wavelength (in meters) of photons with the following energies:

a) 92.0 kJ/mol

b) 8.258 x 10^{-4}kJ/mol

c) 1870 kJ/mol

**Solution:**

1) Determine energy of an individual photon:

92,000 J/mol divided by 6.022 x 10^{23}mol^{-1}= 1.5277 x 10^{-19}JNote how J is used, not kJ.

2) Determine frequency of the photon:

E = hν1.5277 x 10

^{-19}J = (6.626 x 10^{-34}J s) (x)x = 2.30566 x 10

^{14}s^{-1}

3) Determine the wavelength:

λν = cComment: this equals 1300 nm. Compare this to visible light, which ranges from about 400 to 700 nm. If we were to consult a convenient reference source, we would see that our example lies in the infrared region of the EM spectrum.(x) (2.30566 x 10

^{14}s^{-1}) = 3.00 x 10^{8}m/sx = 1.30 x 10

^{-6}m

If you wished, you could use Eλ = hc and solve for the wavelength in one step.

**Problem #4:** A certain green light has a wavelength of 6.26 x 10^{14} Hz. (a) What is its wavelength? (b) What is the energy of one photon of this light? (c) Of one mole of photons?

**Solution:**

1) Use λν = c for (a).2) Use E = hν for (b). Note that you do not need the answer to (a) to solve part (b). This answer is nornally given in units of J.

3) Multiply the answer to (b) by Avogadro's Number. This answer is normally given in kJ/mol.

**Problem #5:** What is the energy of a single UV photon having a wavelength of 11.3 nm? What is the energy of a mole of UV photons having a wavelength of 11.3 nm?

**Solution:**

1) First question: use E = hν. Make sure to convert nm to m first. (You'll use one of these: 1 m = 10^{9}nm or 1 nm = 10^{-9}m)2) Second question: Use Avogadro's Number.

**Problem #6a:** Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is 890.1 kJ/mol. Calculate the minimum wavelength of light that will ionize gold.

**Solution:**

1) We need the energy of one photon:

890100 J/mol divided by 6.022 x 10^{23}mol^{-1}1.478 x 10

^{-18}J

2) This time, use the one step equation:

Eλ = hcλ = hc / E

x = [(6.626 x 10

^{-34}J s) (3.00 x 10^{8}m/s)] / 1.478 x 10^{-18}Jx = 1.345 x 10

^{-7}mYou may wish to satisfy yourself that this is ultraviolet radiation. Remember that the visible spectrum spans 400 to 700 nm.

**Problem #6b:** The ionization energy of gold is 890.1 kJ/mol. Is light with a wavelength of 240. nm capable of ionizing a gold atom (removing an electron) in the gas phase?

**Solution:**

1) Determine the energy required for one photon to ionize one electron:

890100 J/mol divided by 6.022 x 10^{23}mol^{-1}1.478 x 10

^{-18}J

2) Determine the wavelength:

Eλ = hcλ = hc / E

x = [(6.626 x 10

^{-34}J s) (3.00 x 10^{8}m/s)] / 1.478 x 10^{-18}Jx = 1.345 x 10

^{-7}m

3) Convert to nm:

1.345 x 10^{-7}m = 134.5 x 10^{-9}m = 134.5 nm

4) The answer to the question?

No.

**Problem #7:** What is the energy per photon of the lowest frequency of electromagnetic radiation that can be used to observe a gold atom with a diameter of 280. picometers?

**Solution:**

1) Convert 280. pm to m:

280. pm = 280. x 10^{-12}m = 2.80 x 10^{-10}mTo observe an object, we need the wavelength to be as small (or smaller) than the object being viewed. Hence, we need a wavelength of 280. pm.

2) Determine the frequency:

λν = c (2.80 x 10^{-10}m) (x) = 3.00 x 10^{8}m/s x = 1.07143 x 10^{18}s^{-1}

3) Determine the energy:

E = hνx = (6.626 x 10

^{-34}J s) (1.07143 x 10^{18}s^{-1})x = 7.10 x 10

^{-16}J

Comment: this photon lies in the x-ray region of the EM spectrum. And, yes, it is possible to image individual atoms. However, it is not easy.

**Problem #8:** Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs:

AgCl ---> Ag + Cl

The silver metal that is formed causes the lenses to darken. The enthalpy change for this reaction is 310. kJ/mol. Assuming all this energy must be supplied by light, what is the maximum wavelength of light (in nanometers) that can cause this reaction? (Assume 1 photon of light per 1 molecule of AgCl.)

**Solution:**

1) Determine energy of one photon:

310,000 J/mol divided by 6.022 x 10^{23}mol^{-1}5.1478 x 10

^{-19}J

2) Use the one step equation to determine the wavelength:

Eλ = hcλ = hc / E

x = [(6.626 x 10

^{-34}J s) (3.00 x 10^{8}m/s)] / 5.1478 x 10^{-19}Jx = 3.86 x 10

^{-7}m

3) Write the wavelength in nm:

3.86 x 10^{-7}m = 386 x 10^{-9}m = 386 nmFYI, this is UV light.

Comment: the ChemTeam has used photogray lenses since approximately 1972. He loves them! Not as much as his wife or his cats, but fairly close!

**Problem #9:** It takes 208.4 kJ of energy to remove 1 mole of electrons from 1 mole of atoms on the surface of rubidium metal. (a) How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? (b) What is the maximum wavelength of light (in nanometers) capable of doing this?

**Solution:**

1) Determine energy of one photon:

Make sure to use J rather than kJ.

Use Avogadro's Number.

Hint: these energies often have the magnitude of 10^{-19}J.

2) Determine wavelength one of two ways:

Method One: use Eλ = hcMethod Two: use E = hν to get the frequency, then use λν = c to get the wavelength.

Reminder: the equation in method one is simply the result of combining the two equations in method two. I discussed the derivation in Part Two of Light Equations. The method on equation does not have a standard name or a standard form.

3) Convert your answer (which is in meters) to nm. Here are two more problems of this type:

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of lead is 715.0 kJ/mol. Calculate the maximum wavelength of light that will ionize lead.

The ionization energy of aluminum is 577.6kJ/mol. What is the longest wavelength of light for which a single photon could ionize an aluminum atom?

**Problem #10:** Carbon atoms emit photons at 574 nm when exposed to blackbody radiation. How much energy would be obtained if 4.50 moles of propane (C_{3}H_{8}) were pyrolyzed and the resulting carbon atoms exposed to blackbody radiation?

**Solution:**

1) Calculate energy emitted by one carbon atom:

a) convert 574 nm to m

b) Use one of two methods to get the energy:Method One: use Eλ = hcMethod Two: use λν = c to get the frequency, then use E = hν to get the energy.

2) Calculate the number of carbon atoms in 4.50 mol of C_{3}H_{8}:

a) 4.50 mol times Avogadro's Number

b) that answer times three.

3) The final step:

energy emitted by one carbon atom (from step one) times total number of carbon atoms (from step two).2.82 x 10

^{6}J

Wavelength-Frequency-Energy Problems #11 - 20 | Go to Part Two of Light Equations |

Return to Part One of Light Equations | Return to Electrons in Atoms menu |