Problem 9
A nitrogen nucleus (mass $\cong 14\times$ proton mass) emits a photon of energy 6.2 MeV. If the nucleus is initially at rest, what is the recoil energy of the nucleus in eV?
Solution:
With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon. We therefore have its magnitude given by $p=h\nu/\lambda$, where $h\nu=6.2~\mbox{MeV}$. The recoil energy is
$\begin{aligned}
E&=\frac{p^2}{2M}=h\nu\frac{h\nu}{2Mc^2}=(6.2~\mbox{MeV})\frac{6.2~\mbox{MeV}}{2\times 14\times (940~\mbox{MeV})}\\
&=1.5\times 10^{-13}~\mbox{MeV}
\end{aligned}$
Problem 10
Consider a crystal with planar spacing 0.32 nm. What order of magnitude of energies would one need for (a) electrons, (b) helium nuclei (mass $\cong 4\times$ proton mass) to observe up to three interference maxima?
Solution:
The formula $\lambda=2a\sin\theta /n$ implies that $\lambda/\sin\theta\leq 2a/3$. Since $\lambda=h/p$ this leads to $p=\geq 3h/2a\sin\theta$, which implies that the kinetic energy obeys
\[K=\frac{p^2}{2m}\geq \frac{9h^2}{8ma^2\sin^2\theta}\]
Thus the minimum energy for electrons is
\[K=\frac{9(6.63\times 10^{-34}\mbox{Js})^2}{8(0.9\times 10^{-30}\mbox{kg})(0.32\times 10^{-9}\mbox{m})^2(1.6\times 10^{-19}\mbox{J/eV})}=3.35~\mbox{eV}\]
For Helium atoms the mass is $4(1.67\times 10^{-27}\mbox{kg})/(0.9\times 10^{-30}\mbox{kg})=7.42\times 10^3$ larger, so that
\[K=\frac{33.5~\mbox{eV}}{7.42\times 10^3}=4.5\times 10^{-3}~\mbox{eV}\]
Problem 11
The smallest separation resolvable by microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of (a) 15 nm, (b) 0.5 nm?
Solution:
We use $$K=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}$$ with $\lambda=15\times 10^{-9}~\mbox{m}$ to get
\[K=\frac{(6.63\times 10^{-34}~\mbox{Js})^2}{2(0.9\times 10^{-30}~\mbox{kg})(15\times 10^{-9}~\mbox{m})^2(1.6\times 10^{-19}~\mbox{J/eV})}=6.78\times 10^{-3}~\mbox{eV}\]
For $\lambda=0.5~\mbox{nm}$, the wavelength is 30 times smaller, so that the energy is 900 times larger. Thus $K =6.10~\mbox{eV}$.
Problem 12
If one assumes that in a stationary state of the hydrogen atom the electron fits into a circular orbit with an integral number of wavelengths, one can reproduce the results of the Bohr theory. Work this out.
Solution:
For a circular orbit of radius $r$, the circumference is $2\pi r$. If $n$ wavelength $\lambda$ are to fit into the orbit, we must have $2\pi r=n\lambda=nh/p$. We therefore get the condition
\[pr=n\frac{h}{2\pi}=n\hbar\]
which is just the condition that the angular momentum in a circular orbit is an integer in units of $\hbar$.
Problem 13
The distance between adjacent planes in a crystal are to be measured. If X rays of wavelength $0.5~\AA$ are detected at an angle of $5^o$, what is the spacing? At what angle will the second maximum occur?
Solution:
We have $a=n\lambda/2\sin\theta$. For $n=1~,\lambda=0.5\times 10^{-10}~\mbox{m}$ and $\theta=5^o$. We get $a=2.87\times 10^{-10}~\mbox{m}$. For $n=2$, we require $\sin\theta_2 = 2\sin\theta_1$. Since the angles are very small, $\theta_2 = 2\theta_1$. So that the angle is $10^o$.
Monday, November 30, 2009
Saturday, November 28, 2009
Chapter 1 - Problem 6, 7, 8
Problem 6
A 100-keV photon collides with an electron at rest. It is scattered through $90^o$. What is its energy after the collision? What is the kinetic energy in eV of the electron after the collision, and what is the direction of its recoil?
Solution:
Let $h\nu$ be the incident photon energy, the final photon energy and p the outgoing electron momentum. Energy conservation reads
\[h\nu+mc^2=h\nu^\prime+\sqrt{p^2c^2+m^2c^4}\]
We write the equation for momentum conservation, assuming that the initial photon moves in the x -direction and the final photon in the y-direction. When multiplied by $c$ it read
\[\mathbf {i}(h\nu)=\mathbf{ j}(h\nu^\prime)+(\mathbf {i}p_xc+{\mathbf j}p_yc)\]
Hence $p_xc=h\nu~;p_yc=-h\nu^\prime$. We use this to rewrite the energy conservation equation as follows:
\[(h\nu+mc^2-h\nu^\prime)^2=m^2c^4+c^2(p_x^2+p_y^2)=m^2c^4+(h\nu)^2+(h\nu^\prime)^2\]
From this we get
\[h\nu^\prime=h\nu\left(\frac{mc^2}{h\nu+mc^2}\right)\]
We may use this to calculate the kinetic energy of the electron
$\begin{aligned}
K&=h\nu-h\nu^\prime=h\nu\left(1-\frac{mc^2}{h\nu+mc^2}\right)=h\nu\frac{h\nu}{h\nu+mc^2}\\
&=\frac{(100~\mbox{keV})^2}{100~\mbox{keV}+510~\mbox{keV}}=16.4~\mbox{keV}
\end{aligned}$
Also
\[\mathbf{p}c=\mathbf{ i}(100~\mbox{keV})+\mathbf{ i}(-83.6~\mbox{keV})\]
which gives the direction of the recoiling electron.
Problem 7
An electron of energy 100 MeV collides with a photon of wavelength $3\times 10^6~\mbox{nm}$ (corresponding to universal background of blackbody radiation). What is the maximum energy loss suffered by the electron?
Solution:
The photon energy is
$\begin{aligned}
h\nu&=\frac{hc}{\lambda}=\frac{(6.63\times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{3\times 10^6 \times 10^{-9}~\mbox{m}}=6.63\times 10^{-17}~\mbox{J}\\
&=\frac{6.63\times 10^{-17}~\mbox{J}}{1.60\times 10^{-19}~\mbox{J/eV}}=4.14\times 10^{-4}~\mbox{MeV}
\end{aligned}$
The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads
\[\left(\frac{h\nu}{c}\right)^2+p^2+2\left(\frac{h\nu}{c}\right)p\eta_i=\left(\frac{h\nu^\prime}{c}\right)^2+p^\prime^2+2\left(\frac{h\nu^\prime}{c}\right)p^\prime\eta_f\]
Here $eta_i=\pm 1$, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for $\eta_f$. When this is multiplied by $c^2$ we get
\[(h\nu)^2+(pc)^2+2(h\nu)pc\eta_i=(h\nu^\prime)^2+(p^\prime c)^2+2(h\nu^\prime)p^\prime c\eta_f\]
The square of the energy conservation equation, with E expressed in terms of momentum and mass reads
\[(h\nu)^2+(pc)^2+m^2c^4+2Eh\nu=(h\nu^\prime)^2+(p^\prime c)^2+m^2c^4+2E^\primeh\nu^\prime\]
After we cancel the mass terms and subtracting, we get
\[h\nu(E-\eta_i pc)=h\nu^\prime(E^\prime-\eta_f p^\prime c)\]
From this can calculate $h\nu^\prime$ and rewrite the energy conservation law in the form
\[E-E^\prime=h\nu\left(\frac{E-\eta_i pc}{E^\prime-\eta_f p^\prime c}-1\right)\]
The energy loss is largest if $\eta_i=-1~;\eta_f=1$. Assuming that the final electron momentum is not very close to zero, we can write $E+pc=2E$ and $$E^\prime-p^\prime c=\frac{(mc^2)^2}{2E^\prime}$$ so that
\[E-E^\prime=h\nu\left(\frac{2E\times 2E^\prime}{(mc^2)^2}\right)\]
It follows that $$\frac{1}{E^\prime}=\frac{1}{E}+16h\nu$$ with everything expressed in MeV. This leads to $E^\prime = (100/1.64)=61~\mbox{MeV}$ and the energy loss is 39 MeV.
Problem 8
A beam of X rays is scattered by electrons at rest. What is the energy of the X rays if the wavelength of the X rays scattered at $60^o$ to the beam axis is $0.035 \AA$?
Solution:
We have $\lambda^\prime = 0.035\times 10^{-10}~\mbox{m}$, to be inserted into
$\begin{aligned}\lambda^\prime -\lambda&=\frac{h}{m_ec}(1-\cos 60^o)=\frac{h}{2m_ec}=\frac{6.63\times 10^{-34}~\mbox{Js}}{2\times (0.9\times 10^{-30}~\mbox{kg})(3\times 10^{8}~\mbox{m/s})}\\&=1.23\times 10^{-12}~\mbox{m}\end{aligned}$
Therefore
\[\lambda=(3.50-1.23)\times 10^{-12}~\mbox{m}=2.3\times 10^{-12}~\mbox{m}.\]
The energy of the X-ray photon is therefore
\[h\nu=\frac{hc}{\lambda}=\frac{(6.63 \times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{(2.3\times 10^{-12}~\mbox{m})(1.6\times 10^{-19}~\mbox{J/eV})}=5.4\times 10^{5}~\mbox{eV}.\]
A 100-keV photon collides with an electron at rest. It is scattered through $90^o$. What is its energy after the collision? What is the kinetic energy in eV of the electron after the collision, and what is the direction of its recoil?
Solution:
Let $h\nu$ be the incident photon energy, the final photon energy and p the outgoing electron momentum. Energy conservation reads
\[h\nu+mc^2=h\nu^\prime+\sqrt{p^2c^2+m^2c^4}\]
We write the equation for momentum conservation, assuming that the initial photon moves in the x -direction and the final photon in the y-direction. When multiplied by $c$ it read
\[\mathbf {i}(h\nu)=\mathbf{ j}(h\nu^\prime)+(\mathbf {i}p_xc+{\mathbf j}p_yc)\]
Hence $p_xc=h\nu~;p_yc=-h\nu^\prime$. We use this to rewrite the energy conservation equation as follows:
\[(h\nu+mc^2-h\nu^\prime)^2=m^2c^4+c^2(p_x^2+p_y^2)=m^2c^4+(h\nu)^2+(h\nu^\prime)^2\]
From this we get
\[h\nu^\prime=h\nu\left(\frac{mc^2}{h\nu+mc^2}\right)\]
We may use this to calculate the kinetic energy of the electron
$\begin{aligned}
K&=h\nu-h\nu^\prime=h\nu\left(1-\frac{mc^2}{h\nu+mc^2}\right)=h\nu\frac{h\nu}{h\nu+mc^2}\\
&=\frac{(100~\mbox{keV})^2}{100~\mbox{keV}+510~\mbox{keV}}=16.4~\mbox{keV}
\end{aligned}$
Also
\[\mathbf{p}c=\mathbf{ i}(100~\mbox{keV})+\mathbf{ i}(-83.6~\mbox{keV})\]
which gives the direction of the recoiling electron.
Problem 7
An electron of energy 100 MeV collides with a photon of wavelength $3\times 10^6~\mbox{nm}$ (corresponding to universal background of blackbody radiation). What is the maximum energy loss suffered by the electron?
Solution:
The photon energy is
$\begin{aligned}
h\nu&=\frac{hc}{\lambda}=\frac{(6.63\times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{3\times 10^6 \times 10^{-9}~\mbox{m}}=6.63\times 10^{-17}~\mbox{J}\\
&=\frac{6.63\times 10^{-17}~\mbox{J}}{1.60\times 10^{-19}~\mbox{J/eV}}=4.14\times 10^{-4}~\mbox{MeV}
\end{aligned}$
The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads
\[\left(\frac{h\nu}{c}\right)^2+p^2+2\left(\frac{h\nu}{c}\right)p\eta_i=\left(\frac{h\nu^\prime}{c}\right)^2+p^\prime^2+2\left(\frac{h\nu^\prime}{c}\right)p^\prime\eta_f\]
Here $eta_i=\pm 1$, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for $\eta_f$. When this is multiplied by $c^2$ we get
\[(h\nu)^2+(pc)^2+2(h\nu)pc\eta_i=(h\nu^\prime)^2+(p^\prime c)^2+2(h\nu^\prime)p^\prime c\eta_f\]
The square of the energy conservation equation, with E expressed in terms of momentum and mass reads
\[(h\nu)^2+(pc)^2+m^2c^4+2Eh\nu=(h\nu^\prime)^2+(p^\prime c)^2+m^2c^4+2E^\primeh\nu^\prime\]
After we cancel the mass terms and subtracting, we get
\[h\nu(E-\eta_i pc)=h\nu^\prime(E^\prime-\eta_f p^\prime c)\]
From this can calculate $h\nu^\prime$ and rewrite the energy conservation law in the form
\[E-E^\prime=h\nu\left(\frac{E-\eta_i pc}{E^\prime-\eta_f p^\prime c}-1\right)\]
The energy loss is largest if $\eta_i=-1~;\eta_f=1$. Assuming that the final electron momentum is not very close to zero, we can write $E+pc=2E$ and $$E^\prime-p^\prime c=\frac{(mc^2)^2}{2E^\prime}$$ so that
\[E-E^\prime=h\nu\left(\frac{2E\times 2E^\prime}{(mc^2)^2}\right)\]
It follows that $$\frac{1}{E^\prime}=\frac{1}{E}+16h\nu$$ with everything expressed in MeV. This leads to $E^\prime = (100/1.64)=61~\mbox{MeV}$ and the energy loss is 39 MeV.
Problem 8
A beam of X rays is scattered by electrons at rest. What is the energy of the X rays if the wavelength of the X rays scattered at $60^o$ to the beam axis is $0.035 \AA$?
Solution:
We have $\lambda^\prime = 0.035\times 10^{-10}~\mbox{m}$, to be inserted into
$\begin{aligned}\lambda^\prime -\lambda&=\frac{h}{m_ec}(1-\cos 60^o)=\frac{h}{2m_ec}=\frac{6.63\times 10^{-34}~\mbox{Js}}{2\times (0.9\times 10^{-30}~\mbox{kg})(3\times 10^{8}~\mbox{m/s})}\\&=1.23\times 10^{-12}~\mbox{m}\end{aligned}$
Therefore
\[\lambda=(3.50-1.23)\times 10^{-12}~\mbox{m}=2.3\times 10^{-12}~\mbox{m}.\]
The energy of the X-ray photon is therefore
\[h\nu=\frac{hc}{\lambda}=\frac{(6.63 \times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{(2.3\times 10^{-12}~\mbox{m})(1.6\times 10^{-19}~\mbox{J/eV})}=5.4\times 10^{5}~\mbox{eV}.\]
Friday, November 27, 2009
Chapter 1 - Problem 3, 4, 5
Problem 3
Ultraviolet light of wavelength 350 nm falls on a potassium surface. The maximum energy of the photoelectrons is 1.6 eV. What is the work function of potassium?
Solution:
The relationship is
\[h\nu=K+W\]
where $K$ is the electron kinetic energy and $W$ is the work function. Here
\[h\nu=\frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34}~\mbox{Js})(3 \times 10^{8}~\mbox{m/s})}{350 \times 10^{-9}~\mbox{m}}=5.68 \times 10^{-19}~\mbox{J}=3.55~\mbox{eV}\]
With $K=1.60~\mbox{eV}$, we get $W=1.95~\mbox{eV}$.
Problem 4
The maximum energy of photoelectrons from aluminum is 2.3 eV for radiation of 200 nm and 0.90 eV for radiation of 258 nm. Use these data to calculate Planck's constant and the work function of aluminum.
Solution:
We use
\[\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=K_1-K_2\]
since $W$ cancels. From this we get
$\begin{aligned}h&=\frac{1}{c}\frac{\lambda_1\lambda_2}{\lambda_2-\lambda_1}(K_1-K_2)\\
&=\frac{(200 \times 10^{-9}~\mbox{m})(258 \times 10^{-9}~\mbox{m})}{(3 \times 10^{8}~\mbox{m/s})(58 \times 10^{-9}~\mbox{m})}\times (2.3-0.9)~\mbox{eV}\times (1.60 \times 10^{-19}~\mbox{J/eV})\\
&=6.64 \times 10^{-34}~\mbox{J.s} \end{aligned}$
Problem 5
A 100-MeV photon collides with a proton that is at rest. What is the maximum possible energy loss for the photon?
Solution:
The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be $h\nu$, and the backward-scattered photon energy be $h\nu^\prime$. Let the energy of the recoiling proton be $E$. Then its recoil momentum is obtained from $E=\sqrt{p^2~c^2+m^2~c^4}$. The energy conservation equation reads
\[h\nu+mc^2=h\nu^\prime+E\]
and the momentum conservation equation reads
\[\frac{h\nu}{c}=- \frac{h\nu^\prime}{c} + p \]
that is
\[h\nu=-h\nu^\prime+ pc \]
We get $E+pc-mc^2=2h\nu$ from which it follows that
\[p^2~c^2+m^2~c^4=(2h\nu-pc+mc^2)^2\]
so that
\[pc=\frac{4h^2\nu^2+4h\nu mc^2}{4h\nu+2mc^2}\]
The energy loss for the photon is the kinetic energy of the proton
$K=E-mc^2$. Now $h\nu=100~\mbox{MeV}$ and $mc^2=938~\mbox{MeV}$,
so that
\[pc=182~\mbox{MeV}\]
and
\[E-mc^2=K=17.6~\mbox{MeV}.\]
Ultraviolet light of wavelength 350 nm falls on a potassium surface. The maximum energy of the photoelectrons is 1.6 eV. What is the work function of potassium?
Solution:
The relationship is
\[h\nu=K+W\]
where $K$ is the electron kinetic energy and $W$ is the work function. Here
\[h\nu=\frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34}~\mbox{Js})(3 \times 10^{8}~\mbox{m/s})}{350 \times 10^{-9}~\mbox{m}}=5.68 \times 10^{-19}~\mbox{J}=3.55~\mbox{eV}\]
With $K=1.60~\mbox{eV}$, we get $W=1.95~\mbox{eV}$.
Problem 4
The maximum energy of photoelectrons from aluminum is 2.3 eV for radiation of 200 nm and 0.90 eV for radiation of 258 nm. Use these data to calculate Planck's constant and the work function of aluminum.
Solution:
We use
\[\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=K_1-K_2\]
since $W$ cancels. From this we get
$\begin{aligned}h&=\frac{1}{c}\frac{\lambda_1\lambda_2}{\lambda_2-\lambda_1}(K_1-K_2)\\
&=\frac{(200 \times 10^{-9}~\mbox{m})(258 \times 10^{-9}~\mbox{m})}{(3 \times 10^{8}~\mbox{m/s})(58 \times 10^{-9}~\mbox{m})}\times (2.3-0.9)~\mbox{eV}\times (1.60 \times 10^{-19}~\mbox{J/eV})\\
&=6.64 \times 10^{-34}~\mbox{J.s} \end{aligned}$
Problem 5
A 100-MeV photon collides with a proton that is at rest. What is the maximum possible energy loss for the photon?
Solution:
The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be $h\nu$, and the backward-scattered photon energy be $h\nu^\prime$. Let the energy of the recoiling proton be $E$. Then its recoil momentum is obtained from $E=\sqrt{p^2~c^2+m^2~c^4}$. The energy conservation equation reads
\[h\nu+mc^2=h\nu^\prime+E\]
and the momentum conservation equation reads
\[\frac{h\nu}{c}=- \frac{h\nu^\prime}{c} + p \]
that is
\[h\nu=-h\nu^\prime+ pc \]
We get $E+pc-mc^2=2h\nu$ from which it follows that
\[p^2~c^2+m^2~c^4=(2h\nu-pc+mc^2)^2\]
so that
\[pc=\frac{4h^2\nu^2+4h\nu mc^2}{4h\nu+2mc^2}\]
The energy loss for the photon is the kinetic energy of the proton
$K=E-mc^2$. Now $h\nu=100~\mbox{MeV}$ and $mc^2=938~\mbox{MeV}$,
so that
\[pc=182~\mbox{MeV}\]
and
\[E-mc^2=K=17.6~\mbox{MeV}.\]
Wednesday, November 25, 2009
Chapter 1 - Problem 1, 2
Problem 1
Prove the relation (1-1) between the energy density in a cavity and the emissive power. [Hint: To do so, look at the figure. The shaded volume element is of magnitude $r^2 dr \sin\theta~d\theta~ d\phi = dV$, where $r$ is the distance to the origin (at the aperture of area $dA$), $\theta$ is the angle with the vertical, and $\phi$ is the azimuthal angle about the perpendicular axis through the opening. The energy contained in the volume element is $dV$ multiplied by the energy density. The radiation is isotropic, so that what emerges is given by the solid angle $dA \cos\theta/4\pi r^2$ multiplied by the energy. This is to be integrated over the angles $\phi$ and $\theta$ and, if the flow of radiation in time $\Delta t$ is wanted, over $dr$ from 0 to $c~\Delta t$ - the distance from which the radiation will escape in the given time interval.]
Solution:
The energy contained in a volume $dV$ is
\[U(v,T)dV=U(v,T)r^2 dr sin \theta d\theta d\phi\]
when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area $dA$ is
$\begin{aligned}
dE(v,T)&=\int_0^{c\Delta t} dr\int_0^{\pi/2} d\theta~d\phi U(v,T) \sin\theta \cos\theta \frac{dA}{4\pi}\\
&=\frac{dA}{4\pi} 2\pi c~\Delta t~U(v,T) \int_0^{\pi/2} d\theta~ \sin\theta \cos\theta\\
&=\frac{1}{4} c~\Delta t ~dA~ U(v,T)
\end{aligned}$
By definition of the emissivity, this is equal to $E~\Delta t~ dA$. Hence
\[E(v,T)=\frac{c}{4}U(v,T)\]
Problem 2
Given (1-6), calculate the energy density in a wavelength interval $\Delta\lambda$. Use your expression to calculate the value of $\lambda = \lambda_{\rm max}$, for which this density is maximal. Show that $\lambda_{\rm max}$ is of the form $b/T$, calculate $b$, and use 6000K as an estimate of the sun's surface temperature to calculate $\lambda_{\rm max}$ for solar radiation. [Hint: In calculating $b$ you will need the solution $x$ of the equation $(5-x)=5e^{-x}$.]
Solution
We have
\[w(\lambda,T)=U(v,T)~|dv/d\lambda|=U\left(\frac{c}{\lambda}\right)\frac{c}{\lambda^2}=\frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}\]
This density will be maximal when $dw(\lambda,T)/d\lambda=0$. What we need is
\[\frac{d}{d\lambda}\left(\frac{1}{\lambda^5}\frac{1}{e^{A/\lambda}-1}\right)=\left[-5\frac{1}{\lambda^6}-\frac{1}{\lambda^5}\frac{e^{A/\lambda}}{e^{A/\lambda}-1}\left(-\frac{A}{\lambda^2}\right)}\right]\frac{1}{e^{A/\lambda}-1}=0\]
Where $$A=\frac{hc}{kT}$$. The above implies that with $x=A/\lambda$, we must have
\[5-x=5e^{-x}\]
A solution of this is $x=4.965$ so that
\[\lambda_{\rm max}T=\frac{hc}{4.965k}=2.898\times 10^{-3}~\mbox{m}\]
In example 1.1 we were given an estimate of the sun's surface temperature as 6000 K. From this we get
\[\lambda_{\rm max}^{\rm sun}}=\frac{28.98\times 10^{-4}~\mbox{mK}}{6\times 10^3~\mbox{K}}=4.83 \times 10^{-7}~\mbox{m}=483~\mbox{nm}\]
Prove the relation (1-1) between the energy density in a cavity and the emissive power. [Hint: To do so, look at the figure. The shaded volume element is of magnitude $r^2 dr \sin\theta~d\theta~ d\phi = dV$, where $r$ is the distance to the origin (at the aperture of area $dA$), $\theta$ is the angle with the vertical, and $\phi$ is the azimuthal angle about the perpendicular axis through the opening. The energy contained in the volume element is $dV$ multiplied by the energy density. The radiation is isotropic, so that what emerges is given by the solid angle $dA \cos\theta/4\pi r^2$ multiplied by the energy. This is to be integrated over the angles $\phi$ and $\theta$ and, if the flow of radiation in time $\Delta t$ is wanted, over $dr$ from 0 to $c~\Delta t$ - the distance from which the radiation will escape in the given time interval.]
Solution:
The energy contained in a volume $dV$ is
\[U(v,T)dV=U(v,T)r^2 dr sin \theta d\theta d\phi\]
when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area $dA$ is
$\begin{aligned}
dE(v,T)&=\int_0^{c\Delta t} dr\int_0^{\pi/2} d\theta~d\phi U(v,T) \sin\theta \cos\theta \frac{dA}{4\pi}\\
&=\frac{dA}{4\pi} 2\pi c~\Delta t~U(v,T) \int_0^{\pi/2} d\theta~ \sin\theta \cos\theta\\
&=\frac{1}{4} c~\Delta t ~dA~ U(v,T)
\end{aligned}$
By definition of the emissivity, this is equal to $E~\Delta t~ dA$. Hence
\[E(v,T)=\frac{c}{4}U(v,T)\]
Problem 2
Given (1-6), calculate the energy density in a wavelength interval $\Delta\lambda$. Use your expression to calculate the value of $\lambda = \lambda_{\rm max}$, for which this density is maximal. Show that $\lambda_{\rm max}$ is of the form $b/T$, calculate $b$, and use 6000K as an estimate of the sun's surface temperature to calculate $\lambda_{\rm max}$ for solar radiation. [Hint: In calculating $b$ you will need the solution $x$ of the equation $(5-x)=5e^{-x}$.]
Solution
We have
\[w(\lambda,T)=U(v,T)~|dv/d\lambda|=U\left(\frac{c}{\lambda}\right)\frac{c}{\lambda^2}=\frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}\]
This density will be maximal when $dw(\lambda,T)/d\lambda=0$. What we need is
\[\frac{d}{d\lambda}\left(\frac{1}{\lambda^5}\frac{1}{e^{A/\lambda}-1}\right)=\left[-5\frac{1}{\lambda^6}-\frac{1}{\lambda^5}\frac{e^{A/\lambda}}{e^{A/\lambda}-1}\left(-\frac{A}{\lambda^2}\right)}\right]\frac{1}{e^{A/\lambda}-1}=0\]
Where $$A=\frac{hc}{kT}$$. The above implies that with $x=A/\lambda$, we must have
\[5-x=5e^{-x}\]
A solution of this is $x=4.965$ so that
\[\lambda_{\rm max}T=\frac{hc}{4.965k}=2.898\times 10^{-3}~\mbox{m}\]
In example 1.1 we were given an estimate of the sun's surface temperature as 6000 K. From this we get
\[\lambda_{\rm max}^{\rm sun}}=\frac{28.98\times 10^{-4}~\mbox{mK}}{6\times 10^3~\mbox{K}}=4.83 \times 10^{-7}~\mbox{m}=483~\mbox{nm}\]
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