Ultraviolet light of 0.1 um wavelength is incident upon a metal. Which of the metals listed in table 6.2 will emit electrons in response to the input light? Show your reasoning..
Ultraviolet light of 0.1 um wavelength is incident upon a metal. Which of the metals listed in table 6.2 will emit electrons in response to the input light? Show your reasoning..
Related questions
Question
100%
Ultraviolet light of 0.1 um wavelength is incident upon a metal. Which of the metals listed in table 6.2 will emit electrons in response to the input light? Show your reasoning..

Transcribed Image Text:88
Table 6.2 Work functions
of metals
Metal
Li
Na
K
Cs
Cu
Ag
Au
Mg
Ca
Ba
Al
Cr
Mo
Ta
W
Co
Ni
Pt
Work function (eV)
2.48
2.3
2.2
1.9
4.45
4.46
4.9
3.6
3.2
2.5
4.2
4.6
4.2
4.2
4.5
****
668
4.9
5.3
The free electron theory of metals
term in the denominator. We are left then with some Gaussian functions, whose
integrals between 100 can be found in the better integral tables (you can derive
them for yourself if you are fond of doing integrals). This leads us to
4xmkTx/kTep/2mT dpx-
N(px)dpx=
(6.36)
Substituting eqn (6.36) into (6.30) and assuming that r(px) = r is
independent of px, which is not true but gives a good enough approximation,
the integration can be easily performed, leading to
J = Ap(1-r)T²e/KT
where
4лemk²
h³
(6.37)
Ao =
= 1.2 x 106 Am ² K-².
This is known as the Richardson (Nobel Prize, 1928) equation.
The most important factor in eqn (6.37) is exp(-/kT), which is strongly
dependent both on temperature and on the actual value of the work function.
Take, for example, tungsten (the work functions for a number of metals are
given in Table 6.2), for which 4.5 eV and take T = 2500 K. Then, a
10% change in the work function or temperature changes the emission by a
factor of 8.
The main merit of eqn (6.37) is to show the exponential dependence on
temperature, which is well borne out by experimental results. The actual
numerical values are usually below those predicted by the equation, but this is
not very surprising in view of the many simplifications we had to introduce.
In a real crystal, is a function of temperature, of the surface conditions, and
of the directions of the crystallographic axes, which our simple model did not
take into account.
(6.38)
There is one more thing I would like to discuss, which is really so trivial
that most textbooks do not even bother to mention it. Our analysis was one
for a piece of metal in isolation. The electron current obtained in eqn (6.37)
is the current that would start to flow if the sample were suddenly heated to a
temperature T. But this current would not flow for long because, as electrons
leave the metal, it becomes positively charged, making it more difficult for
further electrons to leave. Thus, our formulae are valid only if we have some
means of replenishing the electrons lost by emission. That is, we need an electric
circuit like the one in Fig. 6.3(a). As soon as an electron is emitted from our
piece of metal, another electron will enter from the circuit. The current flowing
can be measured by an ammeter.
A disadvantage of this scheme is that the electrons travelling to the electrode
will be scattered by air; we should really evacuate the place between the emitter
and the receiving electrode, making up the usual cathode-anode configuration
of a vacuum tube. This is denoted in Fig. 6.3(b) by the envelope shown. The
electrons are now free to reach the anode but also free to accumulate in the
vicinity of the cathode. This is bad again because by their negative charge they
will compel many of their fellow electrons to interrupt their planned journey
to the anode and return instead to the emitter. So again we do not measure the
'natural' current.
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
