Extra Credit 27

Last updated
Save as PDF

Page ID: 195753

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

https://phys.libretexts.org/Bookshel...ves_(Exercise)

Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

#62, #95, #106

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)

#18, #41, #54

Solutions

Problem 62.

The wavelengths of visible light range from approximately 400 to 750 nm. What is the corresponding range of photon energies for visible light?

Energy of a photon:

\[E=\dfrac{hc}{\lambda}\]

h=Planck's Constant (6.626 x 10^-34 J s)

c= Speed of Light (3.00 x 10⁸ m/s)

λ= Wavelength

Solve for the wavelengths of 400 and 750 nm:

Recall: 400 nm = 4 x 10^-7 m and 750 nm = 7.5 x 10^-7 m

\[E=\dfrac{(6.626 \times 10^{-34} \hspace{0.5em} J \cdot s)(3.00 \times 10^8 \hspace{0.5em} m/s)}{4 \times 10^-7 \hspace{0.5em} m}\]

= 4.97 x 10^-19 J ⇒ 3.11 eV

\[E=\dfrac{(6.626 \times 10^{-34} \hspace{0.5em} J \cdot s)(3.00 \times 10^8 \hspace{0.5em} m/s)}{7.5 \times 10^-7 \hspace{0.5em} m}\]

= 2.65 x 10^-19 J ⇒ 1.66 eV

Range of photon energies for visible light: 1.66 eV to 3.11 eV or 2.65 x 10^-19 J to 4.97 x 10^-19 J

Problem 95.

When a hydrogen atom is in its third excited state, what are the shortest and longest wavelengths of the photons it can emit?

Upon viewing the energy spectrum of the hydrogen atom the longest and shortest wavelengths are visually distinguishable. 3rd Excited State Hydrogen.png

Using the Rydberg Equation to relate these energy states:

\[\dfrac{1}{\lambda}=R(\dfrac{1}{m^2}-\dfrac{1}{n^2})\]

R= Rydberg Constant (1.097 x 10⁷ m^-1)

m=quantum number for electron in lower energy orbit

n=quantum number for electron in higher energy orbit

Longest wavelength is when n= m+1 = 4 , since the third excited state has a maximum of n=4. Thus m=3 and n=4.

\[\dfrac{1}{\lambda}=1.097\times 10^{7} \hspace{0.5em} m^{-1}(\dfrac{1}{9}-\dfrac{1}{16})\]

=1.88 x 10^-6 m ⇒ 1880 nm

Shortest wavelength is when n → ∞. Thus m=1 and n=4.

\[\dfrac{1}{\lambda}=1.097\times 10^{7} \hspace{0.5em} m^{-1}(\dfrac{1}{1}-\dfrac{1}{16})\]

=9.72 x 10^-8 m ⇒ 97.2 nm

Problem 106.

At what velocity will an electron have a wavelength of 1.00 m?

De Broglie Wavelength:

\[\lambda=\dfrac{h}{mv}\]

h= Planck's constant (6.626 x 10^-34 J s)

m= Mass of electron (9.11 x 10^-31 kg)

v= Velocity of electron

λ= Wavelength

Set the equation equal to the velocity:

\[v=\dfrac{h}{m\lambda}\]

Then solve:

\[v=\dfrac{(6.626 \times 10^{-34} \hspace{0.5em} J \cdot s)}{(9.11 \times 10^{-31} \hspace{0.5em} kg)(1.00 \hspace{0.5em} m)}\]

= 7.27 x 10^-4 m/s

Problem 18.

Which has a greater momentum: an UV photon or an IR photon?

A UV photon would have a greater momentum than the IR photon as the wavelength of the UV is much smaller ranging from 100-400 nm, while that of the IR ranges from 780 nm - 1 mm.

Problem 41.

If an electron and a proton are traveling at the same speed, which one has the shorter de Broglie wavelength?

A proton would have a shorter de Broglie wavelength as the mass of a proton is larger than the mass of an electron.

Problem 54.

Discuss the main difference between an SEM and a TEM.

SEM: Scanning Electron Microscope, relies on analyzing how an electron beam scatters off of the surface of a sample. The sample may be within a few centimeters in size but may require additional preparation depending on the electrical properties the sample holds. Holds a resolving power greater than 1 nm and 250 times the magnification of a light microscope.

TEM: Transmission Electron Microscope, relies on analyzing the transmitted beam that was projected through a sample. The sample must be 100 nm thick, requiring additional preparation of stabilizing biological samples prior to the samples being sliced to size. Holds a resolving power greater than 0.5 Å and 50 million times the magnification.