The equation that relates wavelength, frequency, and speed of light is
#c = lambda*nu#
#c = 3.00xx10^8# #"m/s"# (the speed of light in a vacuum)
#lambda# = wavelength in meters
#nu# = frequency in Hertz (Hz) or #1/"s"# or #"s"^(-1)"#.
So basically the wavelength times the frequency of an electromagnetic wave equals the speed of light.
FYI, #lambda# is the Greek letter lambda , and #nu# is the Greek letter nu (it is not the same as a v).
To find wavelength (#lambda#), the equation is manipulated so that #lambda = c/nu#.
EXAMPLE PROBLEM 1
What is the wavelength of a electromagnetic wave that has a frequency of #4.95xx10^14# #"Hz"#?
Given or Known:
frequency or #nu = 4.95xx10^14# #"Hz"# or #4.50xx10^14# #"s"^(-1)"#
#c = 3.00xx10^8# #"m/s"#
Unknown:
Wavelength or #lambda#
Equation:
#c = lambda*nu#
Solution:
#lambda = c/nu = (3.00xx10^8"m"/color(red)cancel(color(black)("s")))/(4.95xx10^14color(red)cancel(color(black)("s"))^(-1)") = 6.06xx10^(-7)# #"m"#
But what if you don't know the frequency? Can you still find the wavelength? Yes. An equation that relates energy and frequency is:
#E = hnu#
#E# = energy in Joules #("J")#
#h# = Planck's constant = #6.626xx10^(-34)"J"*"s"#
#nu# = frequency = #"Hz"# or #"s"^(-1)"#
To find frequency, the equation is manipulated so that
#nu = E/h#
Once you have frequency, you can use the first equation #c = lambda*nu# to find the wavelength.
EXAMPLE PROBLEM 2
What is the wavelength of an electromagnetic wave having #3.28 xx10^(-19)# #"J"# of energy?
Given or Known:
#E = 3.28xx10^(-19)# #"J"#
#h = 6.626xx10^(-34)# #"J"*"s"#
Unknown:
frequency, #nu#
Equation:
#E = hnu#
Solution: Part 1
#nu = E/h = (3.28xx10^(-19)color(red)cancel(color(black)("J")))/(6.626xx10^(-34)color(red)cancel(color(black)("J"))*"s") = 4.95 xx10^14# #"Hz"# or #4.95xx10^14# #"s"^(-1)#
Solution: Part 2
#lambda = c/nu = (3.00xx10^8"m"/color(red)cancel(color(black)("s")))/(4.95xx10^14color(red)cancel(color(black)("s"))^(-1)") = 6.06xx10^(-7)# #"m"#
