CK-12-Physics - Intermediate

23.2. Photons and the Photoelectric Effect http://www.ck12.org

In the years 1913-1914, Robert A. Millikan (1868-1953) carried out an extensive set of photoelectric experiments
that confirmed Einstein’s predictions.

The photoelectric effect has many practical applications. Automatic doors function by using a photoelectric current.
When a person attempts to enter or leave a doorway, the photoelectric beam is interrupted, causing a drop in the
current that triggers a switch of some sort (perhaps a solenoid, as discussed earlier) which opens the door. When the
beam is no longer interrupted, the circuit is closed, as is the door. The photoelectric effect is used in many common
devices from burglar alarms to smoke detectors to sophisticated electronic circuitry.

In 1921, Albert Einstein (1879-1955) won the Nobel Prize in physics for his 1905 paper on the photoelectric effect.
Interestingly, he never got one for his papers published on the special or general theories of relativity. In 1923,
Robert Millikan won the Nobel Prize in physics for his experimental verification of Einstein’s 1905 paper and for
determining the charge of the electron.

For more information on the photoelectric effect follow the links below.

http://www.youtube.com/watch?v=N7BywkIretM

http://www.youtube.com/watch?v=v5h3h2E4z2Q

Check Your Understanding

1a. What energy do photons of wavelength 400 nm (violet light) have?

Answer:E=h fbutf=cλ, thus

E=

hc

λ

=

( 6. 626 × 10 −^34 J−s)( 3. 00 × 108 ms)

400 × 10 −^9 m

= 4. 968 × 10 −^19 → 1. 97 × 10 −^19 J

1b. Express the answer to part a in electron-volts.

Answer: 1eV= 1. 6 × 10 −^19 J

Therefore,

E=^4.^968 ×^10

− (^19) J

1. 
   

   60 × 10 −^19 eVJ =^3.^11 eV.

   
   


2. 
   

   What is the threshold frequencyfofor copper?

   
   

Answer:Table23.2 gives the work function of copper asWo= 4. 82 eV.

The work function must be expressed in joules, not electron-volts, thus,

Wo= ( 4. 82 eV)

(

1. 60 × 10 −^19 J

eV

)

= 7. 71 × 10 −^19 J→Wo=h fo= 7. 71 × 10 −^19 J→

fo=

Wo

h

=

7. 71 × 10 −^19 J

6. 626 × 10 −^34 J−s

= 1. 16 × 1015 Hz

This frequency corresponds to ultraviolet light (258nm).

Illustrative Example 24.2.1

Assuming the average wavelength of solar radiation is 500 nm and the sun’s power output isP= 3. 83 × 1026 W,
estimate the number of photons the sun emits per second.

Solution: