Chemistry I-Honors
Electromagnetic Spectrum Problems
Solution Set
Constants to know:
8
c = 3.00 x 10 m/sec;
Formulas: c =  ;
Units to remember:
10
12
h = 6.626 x 10 Joule-sec
1 meter = 10 Angstrom = 10 pm
-34
E = h
9
= 10 nm
E = hc/
-1
Units of frequency are in /sec or sec which means there is no unit in the numerator. Frequency is
defined as the number of waves/sec.
1.
What is the frequency of light if it known to have a wavelength of 455 nm?
c = 
 = (3.00 x 1017 nm/sec) / 455 nm = 6.59 x 1014 sec-1
2A. What is the wavelength of light (in nm) if it has a frequency of 2.33 x 10
c = 
17
/sec?
 = (3.00 x 1017 nm/sec) / (2.33 x 1017 /sec) = 1.29
2B. What is this wavelength in Angstroms (A)? in picometers (pm)?
1.29 nm (10 A / 1 nm) = 12.9 A;
1.29 nm ( 1000 pm / 1 nm) = 1.29 x 103 pm
2C. From the chart given, determine what type of electromagnetic radiation it is?
X rays
3.
How much energy is released when an electron drops to the ground state and releases light with a
15
-1
frequency of 5.00 x 10 sec ?
E = h = (6.626 x 10-34 Joule-sec)( 5.00 x 1015 sec-1) = 3.31 x 10-18 J
4.
How much energy (in kJ) is released when one mole of the same type of electron drops back to the
ground state?
3.31 x 10-18 J / 1 photon x (6.02 x 1023 photons / 1 mol) = 1.99 x 106 J
3
= 1.99 x 10 kJ
5.
How much energy (in Joules) is released when an electron releases a photon that has a wavelength of
333 Angstroms?
E = hc /  = (6.626 x 10-34 Joule-sec)( 3.00 x 1018 A/sec) / 333 A
= 5.97 x 10-18 J
6.
-13
If an electron releases one photon with an energy of 3.96 x 10 Joules, what is the wavelength (in cm)
of that wave?
3.96 x 10-13 J = hc /  = (6.626 x 10-34 Joule-sec)( 3.00 x 1010 cm/sec) / 
 = 5.02 x 10-11 cm
7.
If 2.23 moles of photons emit a total of 34.5 kJ of radiant energy, what would be the wavelength (in
Angstroms) of a single photon?
34.5 kJ
1 mole
1000 J
-20
------------- x --------------- x ---------------- = 2.57 x 10 J
2.23 moles 6.02 x 1023
1 kJ
= hc / 
photons
 = 7.73 x 104 Angstroms
8.
What is the energy of the fourth shell? Need to use the Rydberg equation.
En = RH(1/n2) = -2.18 x 10-18 J (1/16) = -1.36 x 10-19 J
9. What is the energy of a photon that is released when an electron drops from the fourth to the third shell?
E = RH (1/n2initial - 1/n2final) = 2.18 x 10-18 J (1/16 - 1/9) = -1.06 x 10-19 J
10. What type of EM radiation is associated with the photon released in #9?
1.06 x 10-19 J = hc/nm
Infrared (IR)