Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2011
Exercise: Light & Energy	© R. Paselk 2008
	Discussion Modules
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Light and Energy

Speed of light in vacuo = c = 2.9979 x 108m/s	c = f•
Plank's constant = h = 6.626 x 10-34J*s	Ephoton = hf = hc/
J = kg m2s-2	= h/mv
	E= mc2

1. What is the frequency of light with a wavelength of 650.4 nm?

From the table above we see that wavelength and frequency are related by the equation:

c = f•

Rearranging we can write

f = c/

But the units of c are in meters, so we will have to convert the wavelength:

(650.4 nm)(1m/10⁹nm) = 650.4 x10^-9m

Plugging into our equations then:

f = (2.9979 x 10⁸m/s)/(650.4 x10^-9m)

f = 4.6093 x 10¹⁴s^-1 = 4.609 x 10¹⁴s^-1

 

2. What is the wavelength of a photon which has a frequency of 90.50 MHz?

From the table above we see that wavelength and frequency are related by the equation:

c = f•

Rearranging we can write

= c/f

Recalling that MHz = 10⁶s^-1, we can plug the numbers into our equation:

= (2.9979 x 10⁸m/s)/(90.50 x 10⁶s^-1)

= 3.3126 m = 3.313 m

3. KHSU broadcasts at 90.50 MHz.

a. Find the energy of one KHSU photon.

From the table we find that energy and frequency (Hz) are related by the equation:

E_photon = hf

Recalling that MHz = 10⁶s^-1 and that h = 6.626 x 10^-34J*s from the table we can plug the numbers into our equation:

E_photon = (6.626 x 10^-34J*s)(90.50 x 10⁶s^-1)

E_photon = 5.9965 x 10^-26 J = 5.996 x 10^-26 J

b. If KHSU has a 9,000 watt transmitter how many photons does it release each second (one watt = one J/sec)

From part (a) we found that one KHSU photon has an energy of 5.9965 x 10^-26 J. Using dimensional analysis we can see that we will have to divide the power of the station (watts) by the energy of the photons:

photons = (power)/(energy/photon) = (9,000 W)/(5.9965 x 10^-26 J/photon)

photons = 1.5009 x 10²⁹ = 2 x 10²⁹

4. Find the wavelength of a ¹²C atom moving at 2 m/sec.

From the table we find that wavelength and mass are related by the equation:

= h/mv

Since a ¹²C atom has a mass of exactly 12 amu and there are 6.022 x 10²³ amu/g we find the mass one atom is:

m = (12 amu)/(6.022 x 10²³ amu/g) = 1.9927 x 10^-23 g

We also know that h = 6.626 x 10^-34J*s. But we need units of mass, so we can use the conversion factor in the table, J = kg m²s^-2 and h = 6.626 x 10^-34 kg m²s^-2*s

= (6.626 x 10^-34 kg m²s^-2*s)/(1.9927 x 10^-23 g)(kg/1,000g)(2 m/s)

= 1.6626 x 10^-23 m = 2 x 10^-8 m

In addition to these exercises you should familiarize yourself with the text materials referenced below.

• Sections 7.1-7.2 (pp 275-284), in your textbook-you should be able to do the approprite examples and exercises in the assigned problems listed on the Schedule.

Light and Energy Module

© R A Paselk

Last modified 11 March 2008