Continued from yesterday:
3. The Photo-Electric Effect:
The famous photo-electric effect is important because of how it highlights the particulate nature of quanta. (So is often also included in previews to quantum theory) While electron diffraction enabled the hypothesis of matter waves or de Broglie waves, the photo-electric effect reinforced the nature of light as photons.
The effect was first observed by Heinrich Hertz in 1887, but it was left for Einstein to explain (and for which he won the Nobel Prize) in 1905. The effect at the time, was most directly observed when a + charged zinc plate (in a Braun type electroscope) was exposed to x-rays or ultraviolet radiation which caused an increased deflection of the electroscope leaf. Conversely, a negatively charged plate exposed to the same high frequency radiation caused a decreased deflection showing a loss of potential. Hertz demonstrated the effect using an apparatus such as shown in Fig. 6.
Fig. 6: Apparatus to investigate the photo-electric effect
To measure the maximum kinetic energy of the photo-electrons one applies a retarding voltage V, gradually increasing it until the most energetic photo-electrons are stopped so the photo-current becomes zero.
At this point:: eVs = K max= ½ mv max2
Thus, the maximum kinetic energy of the electrons can be obtained if Vs is known. If K max is then plotted against the frequency of the incident radiation (for different tests) a graph such as that shown in Fig. 7 is obtained.
Fig. 7: Plotting K max vs. f to get the work function
i)The number of photo-electrons emitted is proportional to the intensity of the incident radiation
ii) The photo-electrons are emitted within an energy range; 0 < K < K max corresponding to a range of frequencies: fo < f < f’. Hence, there exists some frequency (fo) defined as the threshold frequency, below which no electrons are emitted.
From the graph originating in such experiments, it is therefore possible to write:
½ mv max2 = hf - f
Where f is the “work function”. It follows from this that one can also get the following graph in terms of the stopping potential Vs :
Fig. 8: Alternate graph in terms of the stopping potential
And, from experiment: eVs = ½ mvmax2
Therefore, combining the two:
eVs = hf - f
Vs = hf/ e - f/e = (h/e) f - f/e
Where (h/e) is then the slope which can be computed based on the two known quantities (h = 6.626 x 10 -34 Js) and e = 1.6 x 10-19 C. This yields:
(h/e) = (6.62 x 10 -34 Js)/ 1.6 x 10-19 C = 4.13 x 10-15 Js/C
Hence, in such experiments the slope h/e will always remain the same but the y-intercept (f/e) will change. Note that an alternate form of the energy relationship can be written:
½ mvmax2 = hf – (h fo) = h (f - fo) since f = h fo
A beam of radiation consists of bundles of energy of size hf called “photons”. When such photons collide with electrons at or on a metal surface, they transfer an energy hf. The electrons on the metal surface either get all of this energy or none at all. In leaving the surface, electrons lose an amount of energy f which is the work function of the surface. The maximum energy with which an electron can emerge is:
(Energy gained from work function) – (work function)
The fact that K max is independent of the light intensity can be grasped this way: If the light intensity is doubled, the number of photons is doubled which doubles the number of photo-electrons emitted. However, the kinetic energy, which equals hf - f depends only on the frequency of light and the work function, not on the light intensity. Lastly, the fact electrons are emitted almost instantaneously is consistent with the particle theory of light in which the incident energy of light occurs in small packets and there is a one to one interaction between photons and electrons.
Problem: Sodium has a work function of 2.0 eV. Calculate the maximum energy and speed of the emitted electrons when sodium is illuminated by radiation of l = 150 nm. What is the lowest frequency of radiation for which electrons are emitted?
The work function: f = 2 eV = 2(1.6 x 10-19 J)
f = 3.2 x 10-19 J
The incident energy E = hf = hc/l
hc/ l = (6.62 x 10 -34 Js) (3 x 108 ms-1)/ (150 x10-9 m)
hc/ l = 13.2 x 10-19 J
K max = hf - f = [13.2 x 10-19 J - 3.2 x 10-19 J]
K max = 10-18 J
The velocity v = Ö(2 K max /m) =
[(2 x 10-18 J) /(9.1 x 10 -31kg)]1/2 = 1.5 x 10 6 ms-1
Threshold ("cut off") frequency fo = f / h
f / h = (3.2 x 10-19 J)/ (6.62 x 10 -34 Js)
f o = 4.8 x 10 14 Hz
1)When light of l = 0.50 mm falls on a surface it ejects photo-electrons with a minimum velocity of 6 x 10 5 ms-1 Calculate: a) The work function in eV, and b) the threshold frequency for the surface.
2) A stopping potential Vs = 0.54 V is used for photo-electrons dislodged from a metal surface by radiation with l = 750 nm. Find the frequency of the incident radiation and the work function of the metal in electron volts.
3) When light of wavelength 500 nm falls on a surface it produces photo-electrons with a maximum kinetic energy K max = 0.57 eV. Use this data to find the work function in eV and the stopping potential in volts.