 Wavelength
◦ The shortest distance between
equivalent points on a continuous
wave; either crest to crest or trough
to trough
◦ Measured in m
 Frequency
◦ the number of waves that pass a
given point per second; one
hertz (Hz) equals one wave per
second
 Amplitude
◦ The wave's height from the origin to
crest or origin to trough
◦ Measured in m

As shown in the formula, if
wavelength increases, frequency
decreases. As wavelength decreases,
frequency increases.
 speed
c
=λυ
= speed of light
= 3.00 x 108 m/s

If a gamma ray traveling at the speed of
light has a frequency of 2.88 x 1021Hz,
what is its wavelength?
 speed = λ υ
 3.00 x 108 m/s = λ (2.88 x 1021 Hz)
Answer: λ = 1.04 x 10-13 m


Energy will increase as frequency
increases
E=hυ
◦ h is a constant = 6.626 x 10-34 Js

A photon has an energy of 2.93 x 10 -25J.
What is its frequency?
E=hυ
 2.93 x 10-25 J = (6.626x10-34 Js) υ
Answer: υ = 4.42 x 105 Hz or 442,000 Hz

What is the energy of an ultraviolet photon that
has a wavelength of 1.18 x 10-8m?
◦E=hυ
 But wait? Do we have frequency? No? So…Find it!
◦ Speed = c = λ υ
 3.00 x 108m/s = (1.18 x 10-8m) υ
 υ = 2.54 x 1016
◦ Now solve for the energy
 E = (6.626 x 10-34 Js) (2.54 x 1016)
 E = 1.68 x 10-17 J