Extra Credit 31
- Page ID
- 329415
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)
https://phys.libretexts.org/Bookshel...cs_(Exercises)
#29, #41, #56
Question:
A wave function must follow several criteria in order to qualify as a wave function. It must have:
1) Follows the Square Integral and will produce a finite answer
\begin{equation}
\int_{-inf}^{inf} |f(x)^2| \,dx
\end{equation}
2) Wave function is continuous
3) The first derivative of the wave function is continuous
Evaluating each integral (a)-(e)
(a) \begin{equation}
\psi(x)=A*e^{-x^2}
\end{equation}
By evaluating the square integral of this equation, you get the following result:
(\sqrt{\pi}*A^2)/\sqrt{2}
Which has a finite answer suiting condition 1). The original wave function and the first derivative are continuous, thus suiting conditions 2) and 3). Thus, this equation qualifies.
(b) \begin{equation}
\psi(x)=A*e^{-x}
\end{equation}
The square integral for this equation diverges and does not have a finite answer. Thus, it does not follow the first rule and cannot qualify.
(c)\begin{equation}
\psi(x)=A*tan(x)
\end{equation}
The square integral for this wave function also diverges. It is not continuous and will have holes at every value of \pi/2, thus violating rules 1 and 2.
(d)\begin{equation}
\psi(x)=(A*sin(x))/x
\end{equation}
The square integral for this equation creates a finite result of \piA^2, however, the original equation is not continuous at x=0 and thus this does not qualify.
(e)\begin{equation}
\psi(x)=A*e^{-|x|}
\end{equation}
The square integral will converge to \[A^2\] but because of the absolute value, it will only be available in the region of all positive numbers, not all numbers, and thus isn't continuous. The only equation that qualifies is the first equation, (a).
Question:
Schrodinger's Time-Dependent equation is:
\begin{equation}
\frac{\hbar}{2m}* \frac{\partial^2(\psi(x,t))}{\partial x^2}+U(x)\psi(x,t)= i*\hbar*\frac{\partial(\psi(x,t))}{\partial t}
\end{equation}
(a) for the first equation, Asin(kx−ωt)
The first partial derivative
\begin{equation}
\frac{\partial(\psi(x,t))}{\partial x}
\end{equation}
\begin{equation}
= k*A*cos(kx−ωt)
\end{equation}
and the second partial derivative
\begin{equation}
\frac{\partial^2(\psi(x,t))}{\partial x^2}
\end{equation}
\begin{equation}
=-k^2*A*sin(kx−ωt)
\end{equation}
The first partial derivative with respect to t
\begin{equation}
\frac{\partial(\psi(x,t))}{\partial t}
\end{equation}
=\begin{equation}
=-ω*A*cos(kx−ωt)
\end{equation}
Plug the equation into the Schrodinger's Equation:
\begin{equation}
(\frac{-k^2*\hbar^2}{2m}+U(x))*sin(kx−ωt)= -i*\hbar*ω*(cos(kx−ωt))
\end{equation}
There is an imaginary number on the right side of the equation, and because there is no imaginary number is on the left-hand side the two are not equal and cannot satisfy the Time-Dependent Schrodinger Equation
(b) for the second equation, Acos(kx−ωt)
The first partial derivative
\begin{equation}
\frac{\partial(\psi(x,t))}{\partial x}
\end{equation}
\begin{equation}
= -k*A*sin(kx−ωt)
\end{equation}
and the second partial derivative
\begin{equation}
\frac{\partial^2(\psi(x,t))}{\partial x^2}
\end{equation}
\begin{equation}
=k^2*A*cos(kx−ωt)
\end{equation}
The first partial derivative with respect to t
\begin{equation}
\frac{\partial(\psi(x,t))}{\partial t}
\end{equation}
=\begin{equation}
=ω*A*sin(kx−ωt)
\end{equation}
Plug the equation into the Schrodinger's Equation:
\begin{equation}
(\frac{-k^2*\hbar^2}{2m}+U(x))*cos(kx−ωt)= i*\hbar*ω*(sin(kx−ωt))
\end{equation}
For the same reason as in part (a), the imaginary number on one side causes the equation to not satisfy the Time-Dependent Schrodinger Equation.
Question:
The equation of a Particle in a Box is:
\begin{equation}
\frac{h^2}{8\pi^2mL^2}
\end{equation}
where h is Planck constant, 6.626E-34 J*s,
m is the mass of hydrogen, 1.674E-27 kg
L is the width of the box, .2 m
(a) Plugging in each value given above to find the energy of the ground state will provide the following:
\begin{equation}
\frac{6.626E-34 J*s^2}{8\pi^2(2*1.674E-27 kg)*.2m^2}
\end{equation}
or 4.15E-41 Joules. Converting to eV by multiplying 1J/(1.6E-19 eV) = 2.597E-21 eV
(b) We can multiply the equation for the energy of the particle in a box above by some energy level n. By setting this equal to the thermal energy equation:
\begin{equation}
k_b*\frac{T}{2}= E*n^2
\end{equation}
where E is the value calculated from part (a). Solving to find the quantized energy state we find that:
\begin{equation}
n=\sqrt{k_b*\frac{T}{2E}}
\end{equation}
Substituting in for the equation with the Boltzmann constant to be k_b=1.38E-23 J/K and converting Joules to eV, we get:
n=2.24E9 eV for the quantum state n at 300K.
Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)
https://phys.libretexts.org/Bookshel...ves_(Exercise)
#10, #33, #46
#10
Question: In the interpretation of the photoelectric effect, how is it known that an electron does not absorb more than one photon?
Answer: Multiple Photons will not be absorbed by one electron for a couple of reasons. An electron, particularly those in a metal or gas has certain ionization energies. That is to say there is a requisite frequency, or energy, that a photon must have to eject an electron. This property is not additive, so two photons with half the ejection energy can't impact an electron to eject it, only a single photon impact can. The other reason why 2 photos will not eject an electron is that the impact must be elastic in nature, and any reaction between more than one photon will be inelastic.
#33
Question: Suppose an electron in a hydrogen atom makes a transition from the (n+1)th orbit to the nth orbit. Is the wavelength of the emitted photon longer for larger values of n, or for smaller values of n?
Answer: The (n+1) orbital is higher in energy than the n orbital, and because energy is inversely proportional to wavelength via the function:
\(E={h*c}/{\lambda})
This shows that electrons with higher energy levels will have a lower wavelength. For example, an electron in n=5 orbit that goes from the (n+1) to n orbital will produce a short wavelength and inversely release a lot of energy. So, for larger values of n the longer the wavelength emitted will be.
#46
Question: Give an example of an experiment in which light behaves like waves. Give an example of an experiment in which light behaves as a stream of photons.
Answer: One of the first experiments that showed how light behaves like waves was the Double Slit experiment as first performed by Thomas Young in 1801. In this experiment, he shot a beam of light through a panel with two slits and observed the pattern that formed on the plate behind the slits. The plate showed that there were patterns of bright and dark spots. The bright spots were caused by constructive interference of light between the two plates, and the dark spots were created by destructive interference.