Conceptual Physics

this case.)

2. Observe that the electrons that are least bound to the metal will be the electrons that leave it with the maximum kinetic energy. The less
   energy that is “spent” by a photon to free an electron, the more energy there is “left” that can go to increasing the electron’s kinetic
   energy.

Physics principles and equations

Energy is conserved.

The energy of a photon is

E = hf

Step-by-step solution

We converted to electron volts in step six because in quantum physics, energies are commonly stated in electron volts rather than joules. An

electron with kinetic energy KEmax = 1.89 eV would be moving at a speed of 8.19×10^5 m/s, which is approximately 0.27% of the speed of

light. At this speed, relativistic effects do not play a significant role and are ignored in the solution above.

The equation derived above,KEmax = hfíij, was an important result that was used to confirm the particle-like nature of light. Einstein first

stated it in 1905, and Robert A. Millikan experimentally verified it shortly thereafter.

Step Reason

1. E = KEmax + ij conservation of energy

2. KEmax = E – ij solve for KEmax

3. E = hf energy of a photon

4. KEmax = hf – ij substitute equation 3 into equation 2

5. photon energy

6. convert units

7. evaluate

36.7 - Interactive problem: photoelectric effect

Laser engineers use the principles of quantum physics. In the simulation on the

right, you will experiment with one of the fundamental phenomena that underlie the

working of a ruby laser. Rubies are crystals of aluminum oxide with chromium

impurities that give them their distinctive red color. More importantly for engineers,

chromium is primarily responsible for ruby’s lasing properties.

The simulation focuses on one of the outermost electrons of a chromium atom. The

initial step to make a laser work is to excite the electron so that it is in a high energy

state. The electron starts at a low energy state called E 1. Your first goal is to

increase this electron’s energy, and cause it to jump to a higher energy state called

E 3.

How can you boost the energy of the electron from E 1 to E 3? Here, you may fire at

the atom either photons of red light with a frequency of 4.32×10^14 Hz, or photons of

green light with a frequency of 5.45×10^14 Hz. One photon of the red light used in the

simulation has an energy of 1.79 eV, and a photon of the green light has an energy

of 2.26 eV. These photon energies are calculated using the formula E = hf, where

h is Planck’s constant and f is the frequency of the light.

Conduct a few experiments by selecting photons of different colors (energies) and then pressing GO. Do photons of both frequencies have an

effect on the outermost electron of the chromium atom, or just one frequency? After testing the photons, can you deduce what the energy

difference is between E 1 and E 3?

After you fire the appropriate photon, the electron will have energy E 3 , but this energy level is highly unstable. The electron will rapidly and

spontaneously fall to an intermediate energy level E 2. This new state is relatively stable compared to the higher state, and we will pause time in

the simulation when it reaches this “metastable” state. (A photon is emitted when the electron drops from E 3 to E 2 but that photon is irrelevant

to the operation of the laser.)

(^668) Copyright 2007 Kinetic Books Co. Chapter 36