Problem 3.5
Consider the problem of the distribution of blackbody radiation described in Figure 3.3. Note that as T
increases, the wavelength λmax at which u( λ , T) reaches a maximum shifts toward shorter wavelengths.
...
Problem 3.5
Consider the problem of the distribution of blackbody radiation described in Figure 3.3. Note that as T
increases, the wavelength λmax at which u( λ , T) reaches a maximum shifts toward shorter wavelengths.
(a) Show that there is a general relationship between temperature and λmax stating that T × λmax is
constant (Wiens displacement law). (b) Obtain a numerical value for this constant.
Solution
We will begin the problem with Equation (3.9) for the energy density per unit frequency
uf(f; T) = 8πhf3
c3
1
ehf=kBT − 1 (1)
This is equation could be used to sum total energy by integration, R u(f; T)df. In order to obtain the
energy density per unit wavelength, we would want to change the variable of the integration from f to
λ.
uf(f; T)df = uf( c
λ; T)dλ df dλ = uλ(λ; T)dλ (2)
Recall that f = λc . Therefore,
uλ(λ; T) = uf( c
λ; T)dλ df = 8πhc(3λc )3 eh( λc )=k1BT − 1 λc2 = λ5(ehc)8=λk πhc BT − 1) (3)
Note there’s a sign change because the direction of the integration is changed.
In order to get the λmax, we need to find where the derivative of uλ with respect to λ is zero, which
gives us
duλ
dλ =
−5 × 8πhc
λ6(ehc=λkBT − 1) +
8πhcehc=λkBT (hc=λ2kBT)
λ5(ehc=λkBT − 1)2 = 0 (4)
5 λ
=
e
hc
λkBT hc
λ2kBT
e
hc
λkBT − 1
(5)
Letting x = λkhc BT , we should proceed to get
5 =
exx
ex − 1
(6)
1Solving it numerically, we will get x=4.97 is a constant solution of the equation. Therefore at the
extremum λT = kBx
hc is a constant and the numeric value is
λT = hc
kBT =
6:626 × 10−34 × 3 × 108
4:97 × 1:380 × 10−23 m · K = 0:0029m · K (7)
Key Points to remember
- Understanding the conversion from u(f; T ) to u(λ; T ) in EQUATION 3.
2 Problem 3.17
Consider the metals lithium, beryllium, and mercury, which have work functions of 2.3 eV, 3.9 eV, and
4.5 eV, respectively. If light of wavelength 300 nm is incident on each of these metals, determine (a)
which metals exhibit the photoelectric effect and (b) the maximum kinetic energy for the photoelectron
in each case. (c) If the most energetic electrons ejected from the metal are bent into a circular motion
under field B = 4 × 105 T (see the figure below) , what are the radius of the arcs for each metal which
exhibit the photoelectric effect?
Solution
We need to first work out the energy of the incident wave. (Note the value for h is given in the unit of
eV · s and λ is given in the unit of m.)
E = hf = hc
λ =
4:136 × 10−15 × 3 × 108
3 × 10−7 eV = 4:136eV
Therefore we should expect a photoelectric effect only in Lithium and Beryllium where the work function
is lower than the incoming light energy.
The maximum kinetic energy is obtained when there’s no loss in the energy transfer between the photon
and the electron. Then the kinetic energy is the total energy minus the energy required for excitation,
i.e. the work function).
For Lithium, the maximum electron kinetic energy is 4:136eV − 2:3eV = 1:836eV . For Beryllium, the
maximum electron kinetic energy is 4:136eV − 3:9eV = 0:236eV . For Mercury, no electron is excited.
The rest energy of an electron is 0.511MeV which is much greater than our kinetic energy, so we do not
need to consider relativistic effects. Given the kinetic energy we may calculate the velocity of the particle
and hence the radius
mv2
r
= evB (9)
r =
mv
eB
=
q2m · 1 2mv2
eB
(10)
Then for Lithium, the radius of the ejected electron in the field is
p2×0:511MeV=c21:836eV
e×4×10−5T = 0:115m.
Then for Beryllium, the radius of the ejected electron in the field is
p2×0:511MeV=c20:236eV
e×4×10−5T = 0:0409m.
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