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\textsc{University of Wisconsin-Madison, Physics 241 Spring 2014} \\
Grade (entered by Greg Lau):~~~pts/40 pts\\[25pt] % Your university, school and/or department name(s)
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\huge Solutions to problems (Ch. 3)28, 32, 35 and 53;(Ch. 4)2,6,14,15 \\ % The assignment title
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\author{Caitlin Evers, 9068118596} % Your name
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\begin{document}
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% PROBLEM 1
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\section{Problem 3.28}
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Problem statement: Compute the threshold frequency for the photoelectric effect with molybdenum. The work function of molybdenum is 4.22 eV. Discuss whether or not yellow light will cause the ejection of electrons.
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Solution:
We know that at threshold frequency $V_0$ is equal to zero. We can then use $\Phi=hc/\lambda$ to solve for the threshold frequency. Because,
$$eV_0 = hf-\Phi$$
$$0=hf-\Phi$$
$$\Phi=hf$$
And we can substitute $f=\frac{c}{\lambda}$ to get
$$\Phi=\frac{hc}{\lambda}$$
Substituting our numbers into this equation we get,
$$4.22 eV = \frac{1240 eV*nm}{\lambda} $$.
Solving the equation, we find $\lambda=293.84$. Rearranging this equation for frequency,
$$f=\frac {3*10^8}{293.84*10^-9}$$
we find the threshold frequency to be $1.02*10^15$.Yes, yellow light would cause the ejection of photons from molybedenum, because $560nm>294nm$.
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\section{Problem 3.32}
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Problem statement: Compute the work function for cesium.Then calculate the energy of electrons ejected by light of 300nm shining on cesium.
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Solution:
We know that at threshold frequency $\Phi=\frac{hc}{\lambda}$. Using this equation, we can solve for the work function.
$$\Phi = \frac{1240eV*nm}{653nm}$$
We find the work function to be 1.90eV.
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Solving this equation again using 300nm, we find $\Phi$ now to be 4.133eV. The kinetic energy of the ejected electrons is equal to the difference between these two values.
$$4.133-1.90=2.23$$
So the kinetic energy of electrons ejected from cesium due to light of of 300nm is 2.23eV.
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\section {Problem 3.35}
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Problem statement: Find the momentum of a photon when ejected due to wavelengths of (a) 400nm, (b) 0.1nm (c) 3cm, and (d) 2nm.
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Solution: We know that $E=pc$ and that energy of a photon also is $E=hf$. So $E=pc$ $\to$ $p=\frac{E}{c}$
Then,
$$p=\frac {hf}{c}$$.
Substituting $\lambda^{-1}=\frac{f} {c}$ can solve this equation for
$$p= \frac{h}{\lambda}$$
PLugging these numbers in this equation, we find
\begin{itemize}
\item{(a) $1.6565*10^-36$}
\item{(b) $6.626*10^-33$}
\item{(c) $2.2087*10^-41$}
\item{(d) $3.313*10^-34$}
\end{itemize}
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\section {Problem 3.53}
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Problem statement: Determine the fraction of the Sun's energy that is radiated through the visible spectrum.
Solution: The percent of radiation through the visble spectrum can be calculated by taking the integral of Plank's radiation law over the range 350nm-700nm, and dividing it by the integral of Plank's radiation law from 0 to infinity.
$$\frac {\int\limits_{300}^{700}\frac{8\pi hc\lambda^{-5}}{e^{(hc/\lambda kT)}-1} d\lambda}{\int\limits_{0}^{\infty}\frac{8\pi hc\lambda^{-5}}{e^{(hc/\lambda kT)}-1} d\lambda}$$
Plugging this equation in to Mathematica, we find the precentage to be about $43\%$.
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\section {Problem 4.2}
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Problem statement: What transition does the wavelength 379.1nm correspond to using the Balmer equation?
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Solution: We know that Balmer calculated quantum numbers using the equation
$$\lambda_n=364.6(\frac{n^2}{n^{2}-4})$$
We can solve this problem by plugging 379.1 into the left side of the equation.
$$379.1=364.6(\frac{n^2}{n^{2}-4})$$
$$1.0397n^2-4.159=n^2$$
Solving the equation for n, we find the quantum number to be about 10.
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\section {Problem 4.6}
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Problem statement: What fraction of particles with energy 7.0MeV will be scattered at angles greater than 90 degrees in the Rutherford experiment when using gold foil of thickness $2.0\mu$m? What about between the angles of 45 degress and 75 degrees?
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Solution: We know that $f=\pi b^2 nt$.
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We must first find the number of nuclei per unit volume in gold (n). This can be done using Avogadro's number and the atomic mass of gold in the equation $n=\frac{\rho*N_A}{M}$
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So,
$$\frac {(19.3g/cm^3)(6.02*10^23atoms/mol)}{197g/mol} = 5.9*10^{28} atoms/m^3$$
We must also calculate the impact parameter (b) from the equation $b=\frac{k q_{\alpha} Q}{m_{\alpha} v^2}cot{\frac {\theta}{2}}$.
If we ignore $\cot{\frac{\theta}{2}}$ for the moment, we calculate the rest of the equation with
$$\frac {(2)(79)(1.44ev*nm)}{(2)(7*10^6)} = 1.625*10^-5 m^3$$
Then using the above equation with these calculted values,
$$f=(\pi)(1.625*10^-14)(5.9*10^{28})(2*10^-6)*[\cot{\frac{\theta}{2}}]^2$$
We find the answer to be $9.789*10^-5$.
We solve the second part of the quesiton using $f_{45} -f_{75}$.
$$f=(5.705*10^-4)-(1.663*10^-4)$$
We can calculate the radius of a gold atom by multiplying
$$\frac{19.3g}{cm^3} \times \frac{mol}{197g} \times \frac{6.02*10^23}{mol} = (5.8977*10^{22})^{-1}cm^3$$
Setting this value equal to the equation for the volue of a sphere, $V_{sphere}=(4/3)(\pi)(r^3)$ we can solve for radius to be $1.59*10^{-8}$.
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\section {Problem 4.15}
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Problem statement: Calculate the three longest wavelengths in the Lyman series ($n_f=1$) in and indicate their position on a horizontal linear scale.
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Solution: We can use the equation
$$\frac{1}{\lambda}=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$$ for n=1,2,3. We calculate the wavelengths to be 122nm, 103nm, and 87nm respectively.
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\section {Problem 4.16}
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Problem statement: Calculate the quantum number of earth if its angular momentum were quantized like a hydrogen atom. Would the energy release be detectable in a transition to the next lowest level? What would be the radius of that orbit?
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Solution: We know that the mass of Earth is $5.97219*10^24$. We know the radius of Earth is $1.5*10^11$. We can calculate the angular velocity of Earth as follows:
$$v=\frac{2 \pi r}{(365days/year)(24hours/day)(3600sec/hour}= 29885.774m/s$$
Based on the equation
$$n=\frac {mvr}{\hbar}$$
we can calculate the Earth's quantum number as $2.5389*10^74$.
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Because the quantum number is so large, and using the equation
$$r_{n-1}=\frac{\hbar}{mv}(n-1)$$
it is impossible the energy change or the radius of the new orbit.
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