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::

**eV**_{s}= K_{max}= ½ mv_{max}^{2}
Thus,
the maximum kinetic energy of the electrons can be obtained if

**V**is known. If K_{s}_{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*
ii)
The photo-electrons are emitted within an energy range; 0 < K < K

_{max}corresponding to a range of frequencies: f_{o}__<__f__<__f’. Hence, there exists some frequency (f_{o}) 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**

_{max}^{2}**= 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***V**:_{s}_{}

^{}

*Fig. 8: Alternate graph in terms of the stopping potential**empirical graph*based on actual measurements. Recall from the theory:

**½ mv**

_{max}^{2}= hf -**f**

And,
from experiment:

*eV*_{s}= ½ mv_{max}^{2}
Therefore,
combining the two:

eV

**hf - f**_{s}=
or:

V

**hf/ e - f/e = (h/e) f - f/e**_{s}=
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
mv

_{max}^{2}= hf – (h f_{o}) = h (f - f_{o}) since f = h f_{o}**:**

*Einstein’s explanation*
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?

**Solution**:

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 10^{8}ms^{-1})/ (150 x10^{-9}m)
hc/ l = 13.2 x 10

^{-19}J
Therefore:

**K**hf - f = [13.2 x 10

_{max }=^{-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^{-31}kg)]^{1/2}= 1.5 x 10^{6}ms^{-1}
Threshold ("cut off")
frequency f

_{o}= f / h
f / h = (3.2 x 10

^{-19}J)/ (6.62 x 10^{-34}Js)
Therefore:

f

_{o}= 4.8 x 10^{14}Hz**:**

*Problems*
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 V

**= 0.54 V is used**_{s}**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**.**
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