10: Quantum Fundamentals (Exercises)

Q10.6B

Rydberg atoms are excited atoms with more than one electron that possesses a very high principal quantum number. In a Bohr model, the radius of an electron orbit relies on the n value. If n=196, find the energy of the electron in the hydrogen atom.

S10.6B

\[E_n= \dfrac{-13.6eV}{n^2}\]

\[E_n=\dfrac{-13.6eV}{196^2} = 3.54 \times 10^{-4}\,eV\]

Q10.6C

Calculate the wavelength emitted when an electron falls from the n=50 orbit to the n=49 orbit of a hydrogen atom

S10.6C

\[E_n= -\dfrac{1.602\times 10^{-19} Joules}{n^2}\]

\[ \Delta E= -\dfrac{1.602\times 10^{-19}J}{50^2} - -\dfrac{1.602\times 10^{-19}J}{49^2}\]

\[ \Delta E= 2.642\times 10^{-24} Joules\]

\[ \Delta E= E_{photon}= h f ( h= 6.626\times 10^{-34} J*s, f=frequency (1/s)\]

\[ E_{photon} = \dfrac{2.642\times 10^{-24}J}{6.626\times 10^{-34} J*s}= 3.9873\times 10^9 s^{-1} \]

\[ \dfrac{c}{\lambda}=f\]

thus

\[\dfrac{3.00\times 10^8 m/s}{3.9873\times 10^9 s^{-1}}= \lambda=0.0752m\]

\[\lambda=0.0752m\]

Q10.6D

Species that transition at high energy levels, such as \(n=100\), are known as high Rydberg atoms. These are often used in astronomy. Calculate the frequency of light emitted when an electron falls from the \(n=99\) orbit to the \(n=96\) orbit of the hydrogen atom.

S10.6D

We know that the speed of light in vacumm is represented as

\[c=\lambda \nu\]

Additionally, we know that we can solve for the wavelength of an electron transition by

\[\dfrac{1}{\lambda} = R \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right) \]

Combining these equations results in

\[\nu = \dfrac{c}{\lambda} = cR \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)\]

Using the known values, the frequency is represented as

\[\nu = 3\times 10^8 \dfrac{m}{s} \times \dfrac{1.097\times 10^7}{m}* \left(\dfrac{1}{96^2} -\dfrac{1}{99^2}\right)\]

With a final solution of

\[\nu = 2.1314\times 10^{10} Hz\]

Q10.6E

Calculate the wavelength of light emitted when an electron falls from the n = 50 orbit to the n = 49 orbit of the hydrogen atom. Such species are known as high Rydberg atoms. They are detected in astronomy and are more and more studied in the laboratory.

S10.6E

Rydberg equation is given by:

\[\dfrac{1}{\lambda}=R \times (\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\label{1}\]

\[R=1.097 \times 10^7\dfrac{1}{m}\]

Given values in the problem are:

\[n_f = 49\]

\[n_i = 50\]

Plugging into the Rydberg equation results in:

\[\dfrac{1}{\lambda}=1.097 \times 10^7\dfrac{1}{m} \times (\dfrac{1}{49^2}-\dfrac{1}{50^2}) = 180.930\dfrac{1}{m}\]

To find the wavelength, invert the above value. This will give you:

\[\lambda = \dfrac{1}{180.930\dfrac{1}{m}} = .00553m\]

Q10.6F

Calculate the wavelength of light emitted when an electron falls from the n = 100 orbit to the n = 99 orbit of the hydrogen atom. Such species are known as Rydberg atoms. They are detected in astronomy and are more and more studied in the laboratory.

S10.6F

\[\dfrac{1}{\lambda} = R( \dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\]

\[\dfrac{1}{\lambda} = (1.097 \times 10^7)(\dfrac{1}{99^2}-\dfrac{1}{100^2})\]

\[\dfrac{1}{\lambda} = 22.27m^{-1}\]

\[\lambda = \dfrac{1}{22.27m^{-1}} =0.0449m\]

Q10.7

What is the de Broglie wavelength of an nitrogen molecule at room temperature (300K)?

S10.7

The de Broglie wavelength is given by \( \gamma = \dfrac{h}{p} = \dfrac{h}{mv}\)

For N_2, a diatomic molecule, the velocity can be approximated by the root mean square speed with 5 degrees of freedom: thus \(\dfrac{1}{2}mv_{rms}^2 =\dfrac{5}{2}kT\)

\[ \gamma = \dfrac{h}{\sqrt{5mkT}}\]

\[ \gamma = \dfrac {6.626069 \times 10^{-34}\;kg \cdot m^2/s}{\sqrt{5(1.3806 \times 10^{-23}\: m^2kg/s^2\cdot K \times 300 K \times14.0067 amu \times 2\times 1.66054 \times 10^{-27} kg/amu)}}\]

\[\gamma= .02134 nm\]

Q10.8

What is the de Broglie wavelength of a free neutron with kinetic energy of 0.025 eV?

S10.8

Yet again, the de Broglie wavelength is given by \( \gamma = \dfrac{h}{p} = \dfrac{h}{mv}\).

For a free neutron, m_n=1.674929 x 10^-27 kg.

\[KE=\dfrac{1}{2}mv^2\]

\[v=\sqrt{\dfrac{KE\times 2}{ m}}\]

\[\gamma = \dfrac{h}{\sqrt{KE\times 2 m}}\]

\[\gamma = \dfrac{6.626 \times 10^{-34} kg \cdot m^2 /s}{\sqrt{0.025 eV (1.60218 \times 10^{-19} J/eV) \times 2\times 1.674929 \times 10^{-27}kg}}\]

\[\gamma = .1809nm\]

Q10.8A

If the wavelength of a x-ray photon in a x-ray photoelectron spectroscopy (XPS) instrument is 1.25 nm. Calculate the velocity of the electrons emitted from molecules in which the following work functions (e.g., binding energies): 25, 125, 425 eV.

S10.8A

Since \(\nu=(c/\lambda)\) Let's express the formula in the following form:

\[(1/2)m_ev^2=(hc)/\lambda-\phi\]

Where:

• m_e = mass of the electron
• h = Plank's constant

Then:

\[v^2=(2/m_e)(hc/\lambda)-\phi\]

For 25eV:

\(25eV=4.01\times 10^{-18} J\)

We now plug values inside the formula to calculate velocity:

\[v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-4.01\times 10^{-18}J)\]

From there we get the velocity:

\[v=1.84 \times 10^7 m/s\]

For 125eV :

\(125eV=2.00\times 10^{-17} J\)

We plug values inside the formula like before:

\[v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-2.00\times 10^{-17} J)\]

From there we obtain velocity:

\[v=1.75\times 10^7 m/s\]

For 450eV :

\(425eV=6.81\times 10^{-17} J\)

We plug for this values again:

\[v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-6.81\times 10^{-17} J)\]

We obtain velocity:

\[v=1.42\times 10^7 m/s\]

Q10.8B

Photoelectron spectroscopy utilizes the photoelectron effect to measure the binding energy of electrons in molecules and solids, by measuring the kinetic energy of the emitted electrons and using the relation between kinetic energy, wavelength, and binding energy:

\[\dfrac{1}{2}mv^2 = h \nu - \phi\]

One variant of this photoelectron spectroscopy is X-ray photoelectron spectroscopy. If the X-ray wavelength is 1 nm, calculate the velocity of electrons emitted from molecules in which the binding energies are 5, 50, &100 eV

S10.8B

First, solve the equation for velocity:

\[ v = \sqrt{\dfrac{ 2(h \nu - \phi)}{m}} \]

The givens are: $$ \lambda = 1 nm \text{ and } \phi = 5, 50, \text{&} 100 eV\]

There is no wavelength in the equation so we need:

\[ \nu = \dfrac{c}{\lambda} \text{ where } c=2.99 \text{x}10^8 \dfrac{\text{m}}{\text{s}} \text{ (speed of light) } \]

so we get:

\[ v = \sqrt{\dfrac{ 2(\dfrac{hc}{\lambda} - \phi)}{m}} \]

m is the mass of an electron: $$m=9.11\text{x}10^-31 \text{ Kg}\]

and $$hc=1.239\text{x}10^3eV nm\]

Plug in the numbers to the equation:

\[ v = \sqrt{\dfrac{ 2(\dfrac{1.239\text{x}10^3eV nm}{1 nm} - \phi)}{9.11\text{x}10^-31 \text{ Kg}}} \]

The units don't match so we need: $$1eV = 1.602\text{x}10^-19 J\]

So again, plug in the numbers:

\[ v = \sqrt{\dfrac{ 2(\dfrac{1.239\text{x}10^3eV nm}{1 nm} - \phi)*1.602\text{x}10^-19\dfrac{J}{eV}}{9.11\text{x}10^-31 \text{ Kg}}} \]

\[\text{For: } \phi=5 eV, v=2.083\text{x}10^7\]

\[\text{For: } \phi=50 eV, v=2.045\text{x}10^7\]

\[\text{For: } \phi=50 eV, v=2.001\text{x}10^7\]

Q10.8C

Calculate the velocity of electrons emitted from several different materials after the absorption of a photon with a wavelength of 10 nm, given the workfunctions of the materials are 200, 900, and 1000 eV respectively

S10.8C

\[\dfrac{1}{2}mv^2=h\nu-\phi\]

\[v=\sqrt{\dfrac{2(h\nu-\phi)}{m_e}}\]

\[c=\lambda\nu\]

\[\dfrac{c}{\lambda}=\nu\]

\[900pm \dfrac{1m}{1\times 10^{-10}pm}=9\times 10^{-10}m\]

\[200eV \dfrac{1.602\times 10^{-19}J}{1eV}=2.124\times 10^{-17}J\]

\[m_e=9.109\times 10^{-31}kg\]

\[h=6.626\times 10^{-34}J*s\]

\[v=\sqrt{\dfrac{2(\dfrac{(6.626\times 10^{-34}J*s)(2.998\times 10^{8})}{9\times 10^{-10}m}-2.124\times 10^{-17}J)}{9.109\times 10^{-31}kg}}\]

\[v=2.093\times 10^{7} \dfrac{m}{s}\]

\[900eV \dfrac{1.602\times 10^{-19}J}{1eV}=1.442\times 10^{-16}J\]

\[v=\sqrt{\dfrac{2(\dfrac{(6.626\times 10^{-34}J*s)(2.998\times 10^{8})}{9\times 10^{-10}m}-1.442\times 10^{-16}J)}{9.109\times 10^{-31}kg}}\]

\[v=1.296\times 10^{7} \dfrac{m}{s}\]

\[1000eV \dfrac{1.602\times 10^{-19}J}{1eV}=1.602\times 10^{-16}J\]

\[v=\sqrt{\dfrac{2(\dfrac{(6.626\times 10^{-34}J*s)(2.998\times 10^{8})}{9\times 10^{-10}m}-1.602\times 10^{-16}J)}{9.109\times 10^{-31}kg}}\]

\[v=1.153\times 10^{7} \dfrac{m}{s}\]

Q10.8E

Photoelectron spectroscopy utilizes the photoelectron effect to measure the binding energy of electrons in molecules and solids, by measuring the kinetic energy of the emitted electrons and using the relation in problem 10.7 (equation 1) between kinetic energy, wavelength, and binding energy. One variant of this photoelectron spectroscopy is X-ray photoelectron spectroscopy (XPS). If the X-ray wavelength is 0.250 nm, calculate the velocity of electrons emitted from the molecules in which the binding energies are 20, 200, and 2000 eV.

\[\dfrac{1}{2}mv^2 = h\nu-\phi \]

S10.8E

Solving for velocity (v) in equation 1 yields equation 2

\[v=\sqrt{\dfrac{2(h\nu-\phi)}{m_e}} \label{2}\]

where \(h\) is planck's constant, \(\nu\) is the frequency of the X-ray photon, \(\phi\) is the work function or binding energy, and \(m_e\) is the mass of an electron.

Substituting in the relationship between frequency and wavelength yields equation 3

\[v=\sqrt{\dfrac{2[h(\dfrac{c}{\lambda})-\phi]}{m_e}} \label{3}\]

where \(c\) is the speed of light and \(\lambda\) is the wavelength of the emitted electron.

Substituting in the values the binding energy of 20 eV yields

\[v=\sqrt{\dfrac{2[(6.626\times 10^{-34}m^2kg/s)(\dfrac{2.998\times 10^{8}m/s}{0.250\times 10^{-9}m})-20\;eV(\dfrac{1.602\times 10^{-19}J}{1\;eV})]}{9.109\times 10^{-31}kg}}\]

which reduces to

Q10.8F

Photoelectron spectroscopy utilizes the photoelectron effect to measure the binding energy of electrons in molecules and solids, by measuring the kinetic energy of the emitted electrons and using the relation in problem 10.7 between kinetic energy, wavelength, and binding energy. One variant of this is photoelectron spectroscopy is X-ray photoelectron spectroscopy (XPS). If the X-ray wavelength is 0.2 nm, calculate the velocity of electrons emitted from molecules in which the binding energies are 10, 100, and 500 eV.

S10.8F

The relation in problem 10.7: \[ \dfrac{1}{2}mv^2 = h \nu - \phi \]

Solve for velocity, v: \[ v = \sqrt{ \dfrac{2(h \nu - \phi)}{m}} \]

Given: $$ \lambda = 0.2 nm \text{ and } \phi = 10, 100, 500 eV \]

We know that \[ \nu = \dfrac{c}{\lambda} \text{ where } c=2.99E8 \dfrac{m}{s} \]

\[ m_e = 9.11E{-31} \]

\[ 1 eV = 1.602E-19J\]

\[ v = \sqrt{ \dfrac{2 \left( \dfrac{hc}{\lambda} \right) - \phi)*1.602E{-19} \dfrac{J}{eV}}{m}} \]

\[ \text{ when } \phi = 10 \text{ then } v = \sqrt{ \dfrac{2 \left( \dfrac{h*2.99E8 \dfrac{m}{s}}{0.2} \right) - 10)*1.602E{-19} \dfrac{J}{eV}}{9.11E{-31}}} \]

\[ = 4.66E7 \dfrac{m}{s} \]

\[ \text{ when } \phi = 100 \text{ then } v = \sqrt{ \dfrac{2 \left( \dfrac{h*2.99E8 \dfrac{m}{s}}{0.2} \right) - 100)*1.602E{-19} \dfrac{J}{eV}}{9.11E{-31}}} \]

\[ = 4.63E7 \dfrac{m}{s}\]

\[ \text{ when } \phi = 500 \text{ then } v = \sqrt{ \dfrac{2 \left( \dfrac{h*2.99E8 \dfrac{m}{s}}{0.2} \right) - 500)*1.602E{-19} \dfrac{J}{eV}}{9.11E{-31}}} \]

\[ = 4.48E7 \dfrac{m}{s} \]

Q10.9

Calculate the de Broglie wavelengths of the following:

1. A 1-g bullet with velocity 300 m s^-1
2. A 10^-6-g particle with velocity 10^-6 m s^-1
3. A 10^-10-g particle with velocity 10^-10 m s^-1
4. An H₂ molecule with energy of 3/2kT at T = 20 K

S10.9

De Broglie wavelength:

\[λ = \dfrac{h}{mv} = \dfrac{h}{p}\]

where p = momentum, h = Plank's constant (6.626 x 10^-34 Js), v = velocity, m = mass

(a) \[λ = \dfrac{6.626 \times 10^{-34} \space Js}{(1 \times 10^{-3} \space kg)(300 \space m/s)} = 2.21 \times 10^{-33} \space m\]

(b) \[λ = \dfrac{6.626 \times 10^{-34} \space Js}{(10^{-6} \times 10^{-3} \space kg)(10^{-6} \space m/s)} = 6.626 \times 10^{-19} \space m\]

(c) \[λ = \dfrac{6.626 \times 10^{-34} \space Js}{(10^{-10} \times 10^{-3} \space kg)(10^{-10} \space m/s)} = 6.626 \times 10^{-10} \space m\]

(d) k is the Boltzmann constant (1.38 x 10^-23 J/K)

\[E = \dfrac{3}{2}kT = (\dfrac{3}{2})(1.38 \times 10^{-23} \space J/K)(20 \space K)\ = 4.14 \times 10^{-22} \space J\]

\[E = \dfrac{1}{2} mv^2 = \dfrac{p^2}{2m}\]

\[p= \sqrt{2mE}\]

Mass of an H₂ molecule is 3.35 x 10^-27 kg

\[\lambda = \dfrac{6.626 \times 10^{-34} \space Js}{\sqrt{2(3.35 \times 10^{-27} \space kg)(4.14 \times 10^{-22} \space J)}}\ = 3.1 \times 10^{-9} \space m\]

Q10.9A

Protons are accelerated by a 1500 V potential drop.

1. Calculate the kinetic energy of a proton once accelerated.
2. Calculate the de Broglie wavelength of the accelerated proton.

S10.9A

(a) To find kinetic energy we first must calculate the velocity of the accelerated proton. We first find the difference in potential energy using the potential difference and the charge of a proton:

\[\Delta P=P_b-P_a=q(V_b-V_a)=(1.6\times 10^{-19})(-1500)=-2.4\times 10^{-16} J \]

We can find the Kinetic energy by:

\[\Delta K=-\Delta P=+2.4\times 10^{16} J \]

(b) By using the equation for kinetic energy we can rearrange to solve for the velocity:

\[\Delta K = \dfrac{1}{2}mv^2\]

Becomes:

\[v=\sqrt{\dfrac{2\Delta K }{m_p}} = \sqrt{\dfrac{2(2.4\times 10^{16} J) }{1.67\times 10^{-27}}} = 5.36\times 10^5 ms^{-1} \]

Now we solve for the wavelength \(\lambda\) \[\lambda = \dfrac{h}{m_pv} = 2.11\times 10^{-14} cm^{-1}\]

Q10.9B

Electrons are accelerated by a 666-V potential drop.

1. Calculate the de Broglie wavelength.
2. Calculate the wavelength of the X-rays that could be produced when these electrons strike a solid.

S10.9B

Electron charge (C)

\[1.602 \times 10^{-19}\]

Electron mass (kg)

\[9.11 \times 10^{-31}\]

Planck's constant (J s)

\[6.626 \times 10^{-34}\]

Determining the electron's charge

\[E_{electron} = E_{charge} \times V_{drop}\]

\[E_{electron} (J) = E_{charge} \times V_{drop} = 1.066 \times\ 10^{-16}\]

\[E_{electron} = \dfrac{1}{2} mv^{2} = \dfrac{1}{2}\dfrac{p^{2}}{m}\]

Determining wavelength

\[\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}}\]

a. de Broglie wavelength (m)

\[\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}} = 4.75 \times 10^{-11}\]

b. wavelength of X-rays that could be produced when these electrons strike a solid (m)

\[E_{photon} = \dfrac{hc}{\lambda_{photon}} \]

\[\lambda_{photon} = 2.81 \times 10^{-9}\]

Q10.C

A 2000-V potential drop accelerates electrons.

1. Find the de Broglie wavelength.
2. Find the frequency of the X-rays that might be produced when these electrons strike a solid

Potentially useful information:

• charge of an electron \(e=-1.602\times 10^{-16} \;J\)
• mass of an electron \(m_e= 9.109\times 10^{-31} \;kg\)
• Planck's constant \(h=6.626\times 10^{-34} \;Js\)
• speed of light \(c=3\times 10^{8} \dfrac{m}{s}\)

S10.9C

(a) De Broglie's wavelength can be found using the equation:

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

Since the voltage drop is equal to the amount of electrical energy per Coulomb,

\[E=eV=(1.602\times 10^{-16} \;C)(2000 \;V)\]

We also know the electron has kinetic energy, so

\[E=\dfrac{1}{2} m_ev^2\]

We know that,

\[p=m_ev\]

So,

\[E=\dfrac{p^2}{2m_e}\]

Now, rearranging to get an expression for p,

\[p=\sqrt{2m_eE}\]

Substituting this expression for p into the de Broglie wavelength equation,

\[\lambda=\dfrac{h}{\sqrt{2m_eE}}=\dfrac{(6.626\times 10^{-34} \;Js)}{\sqrt{(2)(9.109\times 10^{-31} \;kg)(1.602\times 10^{-16} \;C)(2000 \;V)}} \]

Calculating,

\[\lambda=2.743\times 10^{-11} \;m\]

(b) We can use an equation for energy of a photon in order to find the wavelength of the X-rays.

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

Rearranging,

\[\lambda=\dfrac{hc}{E}=\dfrac{(6.626\times 10^{-34} \;Js)(3\times 10^{8} \dfrac{m}{s})}{(1.602\times 10^{-16} \;C)(2000 \;V)}\]

\[\lambda=6.2041\times 10^{-10} \;m\]

Now that we the X-ray's wavelength, we can find the frequency using

\[c=\lambda\nu\]

Solving for \(\nu\),

\[\nu=\dfrac{c}{\lambda}=\dfrac{(3\times 10^{8} \dfrac{m}{s})}{6.2041\times 10^{-10} \;m}\]

\[\nu=4.836\times 10^{17} \;Hz\]

Q10.9D

Initially stationary electrons are accelerated by a 150-V potential drop.

1. Calculate the De Broglie Wavelength.
2. Calculate the wavelength of the X-Rays that could be produced when these electrons strike a solid.

S10.9D

kinetic energy of an electron is given by

\[m=9.11 \times 10^{-31}kg\]

\[e=1.6 \times 10^{-19}C\]

\[E_{electron}=eV=(1.6 \times 10^{-19})(150) = 2.4 \times 10^{-17} J\]

\[E=\dfrac{1}{2}mv^2\label{1A}\]

Solve for v:

\[v=\sqrt{\dfrac{2KE}{m}}\label{2A}\]

\[v=7.25\times 10^{6}\dfrac{m}{s}\]

\[\lambda=\dfrac{h}{p}=\dfrac{h}{mv}\label{3A}\]

\[\lambda = 1.00\times 10^{-10} m\]

(b)

\[E_{photon} = \dfrac{hc}{\lambda_{photon}}\label{4A}\]

where \(c\) is the speed of light

\[c = 3.00\times 10^{8} \dfrac{m}{s}\]

\[\lambda_{photon} = 8.28*10{-9} m\]

Q9.10

The lifetime of a molecule in a certain electronic state is 10^-10 s. What is the uncertainty in energy of this state? Give the answer in J and in J mol^-1.

S9.10

\[\Delta E \Delta t \geq \dfrac{h}{4 \pi}\]

\[\Delta E \geq \dfrac{h}{4 \pi \Delta t} \geq \dfrac{6.63 \times 10^{-34} \space J \cdot s}{4 \pi (10^{-10} \space s)}\]

\[\Delta E \geq 5.27 \times 10^{-25} \space J\]

\[\Delta E \times (6.022 \times 10^{23} \space 1/mol) \geq 0.318 \space J mol^{-1}\]

Q10.10b

An ultraviolet photon (λ= 120 nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol­­^-1.

S10.10b

Use momentum to find the velocity of the hydrogen molecule:

\[p=\dfrac{h} {λ}\]

\[ λ=120×10^{-9} meters\]

\[h= 6.55×10^{-34}\]\[m_{H_2}=3.32×10^{-27} kg\]

\[p= \dfrac{6.55×10^{-34}Js}{120×10^{-9} meters}=5.46×10^{-27} \dfrac{kg*m}{s}\]

\[v=\dfrac{p}{m_{H_2} }\]

\[\dfrac{5.46×10^{-27}\dfrac{kg*m}{s}}{120×10^{-9} meters}=1.63\dfrac{m}{s}\]

Translational Energy is the same as kinetic energy, so:

\[KE=\dfrac{1}{2}mv^2\]

\[(\dfrac{1}{2})(3.32×10^{-27} kg)(1.63\dfrac{m}{s})^2 = 8.82×10^{-27} Joules\]

In molar terms:

\[(8.82×10^{-27} Joules)(6.022×10^{23} mol^{-1}) = 5.31×10^{-3} \dfrac{Joules}{mol}\]

Q10.11a

1. Find the de Broglie wavelength of N₂ at 450K.
2. How does this relate to the average distance between N₂ molecules in gas phase at 1 atm at 450K?

S10.11a

(a)The wavelength of a particle is given by de Broglie's equation:

\[λ = \dfrac{h}{mv}\]

For N₂,

\[m ≈ 2*14 amu, 1 amu ≈1.66\times 10^{-27} kg \]

v is defined by the relationship between kinetic energy and temperature:

Into de Broglie's equation:

\[λ = \dfrac{6.626\times 10^{-34}*m^{2}*kg*s^{-1}}{(4.648\times 10^{-26}kg)(6.33\times 10^{2}m*s^{-1})}\]

\[λ=2.26\times 10^{-9}m\]

(b)The average distance between molecules is the radius of the volume per molecule, i.e. the full volume divided by the total number of molecules in the sample. Assuming 1 mole, or 6.0221x10²³ molecules of N₂, we can use the Ideal Gas Law to find the radius:

\[PV=nRT, n=1,R=82.058\dfrac{cm^{3}*atm}{K*mol} , P=1\] \[V_{mol}=\dfrac{(82.058\dfrac{cm^{3}*atm}{K*mol})(450K)(1mol)}{1atm}\] \[V_{mol}=3.693\times 10^{4}*cm^{3}\]

\[V_{N_2}=\dfrac{3.693\times 10^{4}*cm^{3}}{6.0221\times 10^{23}}\] \[V_{N_2}=6.13\times 10^{-26}*m^{3}\] \[V=\dfrac{4\pi r^3}{3}\] \[V=\dfrac{4\pi r^3}{3}=6.13\times 10^{-26}*m^{3}\] \[r=2.44\times 10^{-9}m\]

The radius of the volume per molecule begins at the center of each molecule, so we may substract a bond length to the radius to approximate a better answer. The N₂ bond length is of the magnitude of 1x10^-10m, which would reduce our answer only slightly to be 2.34x10^-9m.

The two values are very close given that the calculations of the distance between molecules were rough estimations.

Q10.11C

What is the de Broglie wavelength of a oxygen molecule moving at the average speed at 40 °C? Compare this to the average distance between oxygen molecules in a gas at 1 atm and 40 C.

S10.11C

Start with the equating the kinetic energy to the temperature via the equipartition relationship.

\[\dfrac{1}{2} mv^2=\dfrac{3}{2} kT\]

Solving for velocity, we get:

\[v=\sqrt{\dfrac{3kT}{m}}\]

Now, we need the molecular mass of oxygen (since we are using \(k\)instead or \(R\). Tthe molar mass of \(O_2\) is 32 and to get calculate the molecular mass, we multiply 32 with the mass of a Proton \(1.672\times 10^{27}\):

\[m= (32)(1.672\times 10^{27} kg\]

\[m=5.352\times 10^{-26}\]

This is a single molecule of oxygen so Boltzman constant (k) will be used instead of R the gas constant and 313 K as out temperature.

\[v=\sqrt{\dfrac{3(1.38\times 10^{-23}m^2 kg s^{-1})(313 K)}{5.352\times 10^{-26}kg}}\]

\[v=4.92\times 10^{2} m s^{-1}\]

Then we use the de broglie wave equation \[\lambda=\dfrac{h}{p}\] where momentum (p) is p=mv.

where \(h\) is the planck's constant :

\[\lambda=\dfrac{6.626\times 10^{-34} m^2 kg s^{-1}}{(492 m s^{-1}) (5.352\times 10^{-26} kg)}\]

\[\lambda=2.516\times 10^{-11} m \]

or 25 pm

Assuming that we have 1 cubic meter vessel at 1 barr at 313 K we need to calculate the moles in the vessel per cubic meter.

\[PV=nRT\]

\[n=\dfrac{PV}{RT}\]

\[n=\dfrac{(1 atm)(1 m^3)}{(8.31 JK^{-1} mol^{-1}) (313 K)}\]

\[n= 3.844\times 10^{-4}\ m^3 mol^{-1}\]

Taking the cube root of n will give us n = 0.0727 meters \(mol^{-1}\)

Q10.11D

Which homonuclear diatomic molecule has a de Broglie wavelength of \(12.09 pm\) at room temperature? Assume room temperature is \(25°C\).

S10.11D

Begin with the de Broglie thermal wavelength:

\[\Lambda= \sqrt{\dfrac{h^2}{2\pi mk_BT}}\]

where

h is the Planck constant

m is the mass

\(k_B\) is the Boltzmann constant

T is the temperature.

Also, convert picometers to meters and Celsius to Kelvin before plugging in any values.

To determine the molecule, we will plug in our knowns and solve for the mass(m), not to be confused with the distance in meters:

\[1.209 \times 10^{-11} \;m = \dfrac{6.626 \times 10^{-34}\;J*s}{\sqrt{(2 \pi)(1.38 \times 10^{-23}\;J/K)(m)(298K)}}\]

Now, it is pretty simple algebra to solve for the mass in kg

\[m = 1.1627 \times 10^{-25}\;kg\]

To identify the molecule, let's convert to the more familiar atomic mass units (amu)

\[m = 1.1627 \times 10^{-25}\;kg \times (\dfrac{1 amu}{1.66 \times 10^{-27}\;kg}) ≈ 70.04 amu\]

Remember, this molecule is homonuclear diatomic, so if \(70 amu\) doesn't indicate \(Cl_2\) then dividing by 2 might make it more clear

\[\dfrac{70 amu}{2} = 35 amu \]

Thus, the molecule must consist of two chlorine atoms.

Q10.12A

What is the de Broglie wavelength of a thermal neutron at 350 K? A thermal neutron has the kinetic energy equal to the average kinetic energy of a thermalized monotonic gas (i.e., described by the Maxwell-Boltzmann distribution).

S10.12A

The de Broglie wavelength is determined by

\[\lambda = {\dfrac{h}{p}} = {\dfrac{h}{m \upsilon}}\label{1}\]

where \(p \) is momentum, \(m\) is mass, and \(\upsilon\) is velocity.

A neutron's velocity is determined by

\[\upsilon = (\dfrac{3 k_B T}{m})^{1/2} \label{2}\]

Substituting equation \((2)\) into \((1)\) we get

\[\lambda = \dfrac{h}{(3 m k_B T)^{1/2}}\label{3}\]

where \(h\) is Planck's constant and \(k_B\) is the Boltzmann constant.

We can now solve for our de Broglie wavelength by plugging in \(T=350 K\) and \(m=1.67 \times 10^{-27} kg\) (mass of a neutron)

\[\lambda = \dfrac{6.626 \times 10^{-34} m^2 kg/s}{(3 (1.67 \times 10^{-27} kg) (1.38 \times 10^{-23} J/K) (350 K))^{1/2}} = 1.455 \times 10^{-10} m\]

Q10.12B

What is the (a) temperature and (b) De Broglie wavelength of a thermal proton (proton in thermal equilibrium with its surroundings) with an average Kinetic Energy of 4.14E-16 J?

S10.12B

(a) The average Kinetic Energy (KE) of a proton is related to its temperature by

\[\overline{KE} = \dfrac{3}{2}kT\]

where k is the Boltzmann constant (\(k = 1.38E{-23} m^{2}kg^{1}s^{-2}K^{-1}\)) and \(T\) is in Kelvins.

\[T = \dfrac{2}{3}k\overline{KE}\] \[T = 2.00\times 10^7 K\]

This is one approximation of the temperature at the suns core.

(b) The De Broglie wavelength is given b the following formula

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

where \(h\) is plancks constant (\(h = 6.63E{-34} m^{2}kg^{1}s^{-1}\)), \(p\) is the momentum of the particle, and \(m_p\) is the mass of a proton

\[m_p\overline{KE} = \dfrac{1}{2}p^2\]

\[p = \sqrt{(2m_p\overline{KE})}\]

and therefore

\[\lambda = \dfrac{h}{\sqrt{2m_p\overline{KE}}}\]

Q10.12c

What is the De Broglie wavelength of a thermal neutron at STP? Express your answer in SI units.

S10.22c

A thermal neutron has an average kinetic energy of \[KE=\dfrac{3}{2}kT\] where k is Boltzmann’s constant (1.3806×10⁻²³J/K). The De Broglie wavelength lambda is given by: \[\lambda =\dfrac{h}{p}\] where p is the particle's momentum and h is Planck's constant \(h=6.62607004\cdot 10^{-34}J\cdot s\)

But,

\[p=(2m\times KE)^{\dfrac{1}{2}}\]

For a neutron \(m_{n}=1.67492747\cdot 10^{-27}kg,\)and STP means \(T=273.15K.\) So:

\[p=[2\times(1.67492747\cdot 10^{-27}kg)\times\dfrac{3}{2}\times(1.3806\cdot 10^{-23}J/K)\times273.15K]^{\dfrac{1}{2}}\] \[p=4.3518\cdot 10^{-24}kg\cdot m/s\] \[\lambda =\dfrac{h}{p}=1.5226\cdot 10^{-10}m\]

Q10.13c

Calculate the de Broglie wavelengths of the following:

1. a) A honeybee (0.1 g) flying at 20 mph.
2. b) An electron traveling at 2.00% of the speed of light.

S10.13c

For a honeybee of mass 0.1 g flying at 20 mph, velocity and momentum are

"\[v=\left(\dfrac{20 mi}{1 hr}\right)\left(\dfrac{1610 m}{1 mi}\right)\left(\dfrac{1 hr}{3600 s}\right)=8.94 \dfrac{m}{s}\]

\[p=mv=(0.0001 kg)(8.94\dfrac{m}{s})=8.94\times 10^{-4} kg\dfrac{m}{s}\]

The de Broglie wavelength is "\[\lambda=\dfrac{h}{p}, \text{ where h is } 6.626\times 10^{-34} J.s\]

Therefore, \[\lambda=\dfrac{h}{p}=\dfrac{6.626\times 10^{-34}}{8.94\times 10^{-4}}=7.41\times 10^{-31}m\]

"\[\text{ for an electron with mass of } 9.109\times 10^{-31} kg\text { traveling at 2.00% of the speed of light, velocity and momentum are}\]

\[v=(0.0200)(2.998\times 10^8)=5.99\times 10^6 m.s^{-1}\]

\[p=m_ev=(9.109\times 10^{-31})(8.99\times 10^6)=5.46\times 10^{-24} kg.m.s^{-1}\]

Therefore, \[\lambda=\dfrac{h}{p}=\dfrac{6.626\times 10^{-34}}{5.46\times 10^{-24}}=1.21\times 10^{-10} m=121 pm\]

Q10.13d

Calculate the de Brogile wavelengths of the following :

1. a 1g bullet with velocity 300 m · s^-1
2. a 10^-6 g bullet with velocity 10^-6 m · s^-1
3. a 10^-10g bullet with velocity 10^-10 m · s^-1
4. a H₂ molecule with energy of \[ \dfrac{3}{4}KT \] at T=20 K

S10.13d

(a)

\[1g = 10^{-3}\;kg \]

\[p = mv = 10^{-3}\times300 = 0.3\; kg\cdot m\cdot s^{-1} \]

The de Broglie wavelength is

\[\lambda = \dfrac{h}{p} = \dfrac{6.626 × 10^{-34}}{0.3 } =2.208× 10^{-33}\;kg\cdot m\cdot s^{-1}\]

(b)

\[10^{-6} = 10^{-9}\;kg \]

\[p = mv = 10^{-9}\times10^{-6} = 10^{-15}\; kg\cdot m\cdot s^{-1} \]

The de Broglie wavelength is

\[\lambda = \dfrac{h}{p} = \dfrac{6.626 × 10^{-34}}{10^{-15} } = 6.626 × 10^{-19}\;kg\cdot m\cdot s^{-1}\]

(c)

\[10^{-10} = 10^{-13}\;kg \]

\[p = mv = 10^{-13}\times10^{-10} = 10^{-23}\; kg\cdot m\cdot s^{-1} \]

The de Broglie wavelength is

\[\lambda = \dfrac{h}{p} = \dfrac{6.626 × 10^{-34}}{10^{-23} } = 6.626 × 10^{-11}\;kg\cdot m\cdot s^{-1}\]

(d) The energy of H₂ is \[E=\dfrac{3}{4}K\times 20 =15K\] \[E=\dfrac{p^2}{2m}\] \[p=\sqrt{2mE}\] \[p=\sqrt{30mK}\]

\[\lambda = \dfrac{h}{p} = \dfrac{6.626 × 10^{-34}}{\sqrt{30mK}} = 6.626 × 10^{-34}\times\sqrt{30mK}\;kg\cdot m\cdot s^{-1}\]

Q10.13e

Calculate the de Broglie wavelengths of the following:

1. a .8g bullet with velocity 340ms^-1.
2. a 10^-5g particle with velocity 10^-5ms^-1.
3. a 10^-8g particle with velocity 10^-8ms^-1.
4. an electron moving with velocity 4.8*10⁶ms^-1.

S10.13e

De Broglie wavelength:

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

where P = momentum, h = Plank's constant (6.626*10^-34Js), v = velocity, m = mass

a)

\[\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(.8 \times 10^{-3} \text{Kg})(340 \text{m/s})} = 2.44 \times 10^{-33} \text{m}\]

b)

\[\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(10^{-5} \times 10^{-3} \text{Kg})(10^{-5} \text{m/s})} = 6.626 \times 10^{-21} \text{m}\]

c)

\[\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(10^{-8} \times 10^{-3} \text{Kg})(10^{-8} \text{m/s})} = 6.626 \times 10^{-15} \text{m}\]

b)

\[\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(9.1 \times 10^{-31} \text{Kg})(4.8^{6} \text{m/s})} = 1.52 \times 10^{-10} \text{m}\]

Mass of an electron is 9.1*10^-31 kg.

Q10.13f

What is the de Broglie wavelength for a proton traveling at 1000 times the speed of sound?

S10.13f

The mass of a proton is given by:

\[m{_{p}}=1.672*10^{-27}\]

The speed of sound is given by:

\[v{_{sound}}=340.29 m/s\]

This gives us a value of momentum as follows:

\[p=(v{_{sound}}*1000)*m{_{p}}=340.29*1000*1.672*10^{-27}=5.6896*10^{-22}\]

Planck's constant is:

\[h=6.626*10^{-34}\]

This gives us the de Broglie wavelength as follows:

\[\lambda =\dfrac{h}{p}=\dfrac{6.626*10^{-34}}{5.6896*10^{-22}}=1.1646*10^{-12}m\]

Q9.14

For a particle in a one-dimension box, the ground-state wavelength is

\[\phi = \left(\dfrac{2}{a}\right)^{1/2}sin{\dfrac{\pi x}{a}}\]

1. What is the probability that the particle is in the right-hand half of the box?
2. What is the probability that the particle is in the middle third of the box?

S9.14

(a) The right-hand half of the box is \(\dfrac{L}{2} \leq x \leq L\)

\[ P\left(\dfrac{L}{2} ≤ x ≤ L\right)=\int_{\dfrac{L}{2}}^{L} \psi^*(x) \psi(x)dx \\ =\int_{\dfrac{L}{2}}^{L}\dfrac{2}{L}\sin^2{\dfrac{\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{\dfrac{L}{2}}^{L}(1-\cos{\dfrac{2\pi x}{L}})dx=\\ \dfrac{1}{L}x\Biggr\rvert_{\dfrac{L}{2}}^{L} +\dfrac{1}{L}\dfrac{L}{2\pi}(-\sin{\dfrac{2\pi x}{L}})\Biggr\rvert_{\dfrac{L}{2}}^{L} \\= \dfrac{1}{L}L-\dfrac{1}{L}\dfrac{L}{2}-\dfrac{1}{2\pi}\sin{\dfrac{2\pi L}{L}}+\dfrac{1}{2\pi}\sin{\dfrac{2\pi L}{2L}}=\\ 1-\dfrac{1}{2}-\dfrac{1}{2\pi}\sin{2\pi}+\dfrac{1}{2\pi}\sin{\pi} \\= \dfrac{1}{2} - \dfrac{1}{2n\pi} (0)+\dfrac{1}{2n\pi} 0 = \\ \dfrac{1}{2} - 0 + 0 \\ = \dfrac{1}{2} = 0.50 \]

(b) The middle third of the box is \(\dfrac{L}{3} \leq x \leq \dfrac{2L}{3}\)

\[ P\left(\dfrac{L}{3} ≤ x ≤ \dfrac{2L}{3}\right)\\=\int_{\dfrac{L}{3}}^{\dfrac{2L}{3}} \psi^*(x) \psi(x)dx\\=\int_{\dfrac{L}{3}}^{\dfrac{2L}{3}}\dfrac{2}{L}\sin^2{\dfrac{\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{\dfrac{L}{3}}^{\dfrac{2L}{3}}(1-\cos{\dfrac{2\pi x}{L}})dx=\\ \dfrac{1}{L}x\Biggr\rvert_{\dfrac{L}{3}}^{\dfrac{2L}{3}} +\dfrac{1}{L}\dfrac{L}{2\pi}(-\sin{\dfrac{2\pi x}{L}})\Biggr\rvert_{\dfrac{L}{3}}^{\dfrac{2L}{3}}= \dfrac{1}{L}\dfrac{2L}{3}-\dfrac{1}{L}\dfrac{L}{3}-\dfrac{1}{2\pi}\sin{\dfrac{2\pi 2L}{3L}}+\dfrac{1}{2\pi}\sin{\dfrac{2\pi L}{3L}}=\\ \dfrac{2}{3}-\dfrac{1}{3}-\dfrac{1}{2\pi}\sin{\dfrac{4\pi}{3}}+\dfrac{1}{2\pi}\sin{\dfrac{2\pi}{3}}= \dfrac{1}{3} - \dfrac{1}{2n\pi} (-\dfrac{\sqrt{3}}{2})+\dfrac{1}{2n\pi} \dfrac{\sqrt{3}}{2} = \\ \dfrac{1}{3} + \dfrac{1}{2\pi} \sqrt{3} = \dfrac{1}{3} + 0.28 = 0.61 \]

Q10.15

For a particle in a one-dimensional box, the ground-state wavefunction is

\[\phi = \sqrt{\dfrac{2}{a}} sin(\dfrac{\pi x}{a})\]

1. What is the probability that the particle is in the left quarter of the box?
2. What is the probability that the particle is between 3/8 the length of the box and 5/8 the length of the box?

S10.15

a) The probability that the particle can be found in the first quarter of the box is found using:

\[ prob = \int^{ \dfrac{a}{4} }_0 |\phi|^2\,dx = \int^{\dfrac{a}{4}}_0 \dfrac{2}{a}sin^2 ( \dfrac{\pi x}{a}),dx \]

The solution to this integral requires the trig identity

\[ sin^2 (x) = \dfrac{1}{2}(1 - cos(2x) ) \]

Using this and computing the integral results in the solution:

\[ prob = \dfrac{1}{2} - \dfrac{1}{2\pi} \]

b) The probability that the particle can be found between the 3/8 and 5/8 the length of the box is found in the exact same manner, this time changing the bounds to

\[ \int^{ \dfrac{5}{8}a }_{\dfrac{3}{8}a} |\phi|^2\,dx \]

The solution is:

\[ prob = \dfrac{1}{4} \]

Q10.19

1. Determine the energy levels for for n = 1,2, and 3 in a potential well that is 0.29 nm wide and has infinite barriers on either side.
2. If an electron makes a transition from n = 1 to n = 3, what is the wavelength of emitted radiation.

Q10.17C

What are the results of operating on the following functions with following two operators:

• \( \hat{A} = \dfrac{d}{dx}\) and
• \( \hat{B} = \dfrac{d^2}{dx^2}\)

1. \(3e^{-cx^3}\)
2. \(\cos(4ax^2)\)
3. \(e^{i^2kx^3}\)

S10.17C

a)

1. Apply Operator: \[ \dfrac{d(3e^{-cx^3})}{dx}\]Take derivative of the whole function by multiplying the exponential by the derivative of its superscript: \[ -9cx^{2}e^{-cx^{3}} \]
2. Apply Operator: \[ \dfrac{d^2(3e^{-cx^3})}{dx^2}\]Which is the same as: \[ \dfrac{d(\dfrac{d(3e^{-cx^3})}{dx})}{dx}\]Or: \[ \dfrac{d(-9cx^{2}e^{-cx^{3}})}{dx}\]And take the derivative the same way as before: \[ 27c^{2}x^{4}e^{-cx^{3}} \]

b)

1. Apply Operator: \[ \dfrac{d(\cos(4ax^2))}{dx}\]Take derivative with respect to cosine first, and then with respect to the function within the cosine: \[ -8ax\sin(4ax^{2}) \]
2. Apply Operator: \[ \dfrac{d^2(\cos(4ax^2))}{dx^2}\]Which is the same as: \[ \dfrac{d(\dfrac{d(\cos(4ax^2))}{dx})}{dx}\]Or: \[ \dfrac{d(-8ax\sin(4ax^{2}))}{dx}\] And take the derivative the same way as before, except now you need the Chain Rule. Take the derivative of the 'x' term before the sine and multiply by the sine function. Then take the derivative of the sine function and multiply by the 'x' function: \[ -8a\sin(4x^{2})-64a^{2}x^{2}\cos(4ax^{2}) \]

c)

1. Apply Operator: \[ \dfrac{d(e^{i^2kx^3})}{dx}\]Take derivative of the whole function by multiplying the exponential by the derivative of its superscript. Note that the 'i' within the superscript refers to the imaginary number 'i':\[ -3kx^{2}e^{-kx^{3}} \]
2. Apply Operator: \[ \dfrac{d^2(e^{i^2kx^3})}{dx^2}\]Which is the same as: \[ \dfrac{d(-3kx^{2}e^{-kx^{3}})}{dx}\]And take the derivative the same way as before, except now you need the Chain Rule. Take the derivative of the 'x' term before the exponential and multiply by the exponential function. Then take the derivative of the exponential function and multiply by the 'x' function: \[ -6kxe^{-kx^{3}}+9k^2x^4e^{-kx^3} \]And finally, simplify:\[3kxe^{-kx^{3}}(3kx^3-2)\]

Q10.17D

What are the results of operating on the following functions with the operator

\[\dfrac{d}{dx} \]

and the operator

\[\dfrac{d^2}{dx^2} \]

1. \(\ln{(-ax^2)} \)
2. \(tan({bx}) \)
3. \(e^{(ikx)} \)

Which functions are eigenfunctions of these operators? What are the corresponding eigenvalues?

S10.17D

(a) First operate on the function using the operators given.

\(\dfrac{d}{dx} \) \(\ln{(-ax^2)} \) = \(\dfrac{2}{x}\)

\(\dfrac{d^2}{dx^2} \) \(\ln{(-ax^2)} \) = \(-\dfrac{2}{x^2}\)

(b)

\(\dfrac{d}{dx} \) \(tan({bx}) \) = \(bsec^2({bx}) \)

\(\dfrac{d^2}{dx^2} \) \(tan({bx}) \) = \(b^2sec^2({bx})tan({bx})\)

(c)

\(\dfrac{d}{dx} \) \(e^{(ikx)} \) = \(ike^{(ikx)} \)

\(\dfrac{d^2}{dx^2} \) \(e^{(ikx)} \) = \(-k^2e^{(ikx)} \)

Now to determine if the functions are eigenfunctions of these operators, we take the result after operating on the function, and if that result is a constant times the original function, then the function is a eigenfunction of that operator. In this case, For (a), x is not a constant, so it is not an eigenfunction. For (b), although b is a constant, it is not that constant times a constant for both \(\dfrac{d}{dx} \) and \(\dfrac{d^2}{dx^2} \). For (c), we do indeed have an eigenfunction, \(ik\) is a constant, times its original function \(e^{(ikx)} \). The same goes for \(\dfrac{d^2}{dx^2} \), we have a constant \(-k^2\) times the original function \(e^{(ikx)} \). The eigenvalues are the constants of these functions, \(ik\) and \(-k^2\) for \(\dfrac{d}{dx} \) \(e^{(ikx)} \) and \(\dfrac{d^2}{dx^2} \) \(e^{(ikx)} \) respectfully.

Q10.17E

What are the results of operating on the following functions with the operator \(\dfrac{d}{dx}\) and \(\dfrac{d^2}{dx^2}\) :

1. \(e^\dfrac{-Zx^2}{a_0}\)
2. \(\cosh bk\)
3. \(kx\)

S10.17E

a) \( \dfrac{d}{dx}\left(e^\dfrac{-Zx^2}{a_0}\right) \)

\[ = \dfrac{-2zx}{a_0} e^\dfrac{-Zx^2}{a_0} \]

Since \( \dfrac{-2zx}{a_0}\) isn't constant this isn't an eigenfunction for this operator

\(\dfrac{d^2}{dx^2}\left(e^\dfrac{-Zx^2}{a_0}\right) \)

\[ = \dfrac{d}{dx}\left (\dfrac{-2zx}{a_0} e^\dfrac{-Zx^2}{a_0}\right) \]

\[ = \dfrac{-2zx}{a_0} e^\dfrac{-Zx^2}{a_0} + \dfrac{4z^2x^2}{a^2_0} e^\dfrac{-Zx^2}{a_0} \]

\[ = \left( \dfrac{-2zx}{a_0} + \dfrac{4z^2x^2}{a^2_0}\right) e^\dfrac{-Zx^2}{a_0}\]

Since the \(\left( \dfrac{-2zx}{a_0} + \dfrac{4z^2x^2}{a^2_0}\right) \) isn't constant this isnt an eigenfunction for this operator either

b) \( \dfrac{d}{dx}\left(\cosh bx\right) \)

\[ = b \sinh bx\]

Not an eigenfunction since we didn't get our original function back for this operator

\[ \dfrac{d^2}{dx^2}\left(\cosh bx\right) \]

\[ = \dfrac{d}{dx}\left(b \sinh bx\right) \]

\[ = b^2 \cosh bx \]

For this operator \( b^2 \) is the eigenvalue and \( \cosh bx\) is the eigenfunction

c) \( \dfrac{d}{dx}\left(kx\right) \)

\[ = k = \dfrac{1}{x} kx \]

Since we get a variable term this isnt an eigenfunction for this operator.

\[ \dfrac {d^2}{dx^2}\left(kx\right) = 0 \]

For this operator \(0\) is our eigenvalue and \(kx\) is our eigenfunction.

Q10.19A

1. Calculate the energy levels for n = 1, 3, and 5 for an electron in a potential well of width 0.50 nm with infinite barriers on either side.
2. If an electron makes a transition from n = 3 to n = 1, what will be the wavelength of the emitted radiation?

S10.19A

(a)

Solve by using the following general equation for the expectation value for energy for a particle in a 1D box:

\[E_n = \dfrac{n^2 h^2}{8 m L^2}\]

Solve for n = 1, n = 2, and n = 3 where \(h = 6.626\times 10^{-34} \;J \cdot s\), \(m = 9.109\times 10^{-31} \;kg\), and \(L = 5.0\times 10^{-10} \;m\):

\[E_1 = \dfrac{(1)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 2.41\times 10^{-19} \;J\]

\[E_3 = \dfrac{(3)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 2.17\times 10^{-18} \;J\]

\[E_5 = \dfrac{(5)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 6.02\times 10^{-18} \;J\]

(b)

Find the energy emitted by finding the difference between E₃ and E₁:

\[E_3 - E_1 = 2.17\times 10^{-18} \;J - 2.41\times 10^{-19} \;J = 1.93\times 10^{-18}\;J\]

Find the wavelength of the energy emitted by the Planck-Einstein relation:

\[\Delta E = h\nu\]

\[\Delta E = h(\dfrac{c}{\lambda})\]

\[\lambda = 1.03\times 10^{-7}\;m\]

Q10.20A

For a helium atom in a one-dimensional box calculate the value of the quantum number of the energy level for which the energy is equal to 5/4 kT at 30°C for:

1. a box \(10^{-3}\;m\) long,
2. a box \(10^{-5}\;m\) long, and
3. a box \(10^{-9}\;m\) long.

S10.20A

For parts (a)-(c) we need to rewrite the following equation in terms of n:

\[E_n=\dfrac{n^{2}h^{2}}{8mL^{2}}\]

Solving for n we get:

\[n=\sqrt{\dfrac{E_n8mL^{2}}{h^{2}}}\]

where

\[E_n=\dfrac{5}{4}(1.38\times10^{-23}m^{2}kgs^{-2}K^{-1})(303K)=5.23\times10^{-21}J\]

Plug in each value of L in meters and solve

1. \[n=\sqrt{\dfrac{(5.23\times10^{-21}J)8(6.647\times10^{-27}kg)(10^{-3})^{2}}{(6.67\times10^{-34}J/s)^{2}}}=2.5\times10^{7}\]
2. \[n=\sqrt{\dfrac{(5.23\times10^{-21}J)8(6.647\times10^{-27}kg)(10^{-5})^{2}}{(6.67\times10^{-34}J/s)^{2}}}=2.5\times10^{5}\]
3. \[n=\sqrt{\dfrac{(5.23\times10^{-21}J)8(6.647\times10^{-27}kg)(10^{-9})^{2}}{(6.67\times10^{-34}J/s)^{2}}}=25\]

Q10.20B

For a helium atom in a one-dimensional box calculate the value of the quantum number of the energy level for which the energy is equal to 2 kT at 25ºC.

1. box \(1\; nm\) long
2. box \(10^^{-6}\;m\) long
3. box \(10^{-2}\;m\) long

S10.20B

\[En= \dfrac{n^2·\hbar·pi^2}{2·m·L^2}\ = \dfrac{n^2·h^2}{8·m·L^2}\]

\[En= 2kT\] \[k= Boltzmanns constant\ 1.38\times 10^-23 \dfrac{m^2·kg^2}{s^2 K}\]

a) \[2·1.38·10^-23·298 = \dfrac{n^2·(6.634\times 10^-34)^2}{8·2·(1.66\times 10^-27\times 10^-18)}\]

\[8.22\times 10^-21 = n^2·1.675\times 10^-23\]

\[n^2= 490.746\]

\[n=22.15\]

b) \[2·1.38·10^-23·298 = \dfrac{n^2·(6.634\times 10^-34)^2}{8·2·(1.66\times 10^-27\times 10^-12)}\]

\[8.22\times 10^-21 = n^2·1.675\times 10^-29\]

\[n^2= 490736341\]

\[n=22152.569\]

c)

\[2·1.38·10^-23·298 = \dfrac{n^2·(6.634\times 10^-34)^2}{8·2·(1.66\times 10^-27\times 10^-4}\]

\[8.22\times 10^-21 = n^2·1.675\times 10^-37]

\[n^2= 4.9\times 10^16\]

\[n=221525696.2\]

Q10.20C

For a Helium atom in one-dimensional box calculate the value of the quantum number of the energy level for which the energy is equal to 2KT at 25 ℃

1. for a box 10 nm long,
2. for a box \(2\times 10^{-6} m\) long,
3. for a box \(5\times 10^{-2} m\) long.

S10.20C

T = 25℃ = 298.15 K. K = 1.381\times 10^(-23) JK^(-1). So the energy equals to E = 6.176*10(-21) J.

we know that $$E_n=\dfrac{h^2*n^2}{8*M_e*a^2}$$ where Me is the electron weight and a is the length of the box.

To get the energy level, we use the function and plug in all the data. $$M_e=9.109\times 10^(-31)Kg\]

\[n=\dfrac{a}{h}\sqrt{8*M_e*E_n}\]

Then we could get:

1. n = 0.3202
2. n = 640.3361
3. n = 16008000

Q10.20D

For a helium atom in a one-dimensional box, calculate the quantum number for the wavefunctions with the energies equal to the average kinetic energy of a thermalized monotonic gas (e.g, 3/2 kT) for a box 1 nm long at -100 ^oC, 0 ^oC, and 100 ^oC.

S10.20D

Use Energy Equation:

\[E_{n}=\dfrac{n^{2}h^{2}}{8mL^{2}}\]

Substitute Average Energy from Question for En

\[\dfrac{3}{2} \times kT=\dfrac{n^{2}h^{2}}{8mL^{2}}\]

Solve this equation for n:

\[n=\sqrt{\dfrac{12kTmL^{2}}{h^{2}}}(1)\]

Convert 1 nm to _ m

\[L=10^{-9} m\]

Convert each temperature from ^oC to K

Then plug in the numbers into the Equation (1) to find the quantum number for the different temperatures:

\[T=173K\]

\[n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 173K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}\]

\[n=21\]

\[T=273K\]

\[n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 273K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}\]

\[n=26\]

\[T=373K\]

\[n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 373K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}\]

\[n=31\]

S10.20e

Now, Notice the lowest Energy molecular orbital is effectively equal to the 1s_x orbital. This is a core atomic orbital and is not involved in the bonding. (Recall that bonding orbitals involve the valence shell electrons, which for the X atom are in the orbitals with the highest n quantum number. Which is n XH₃​

Q10.22a

Calculate the degeneracies of the first 3 levels of a particle in a 4D box of length L

S10.22a

\[ E(n_x,n_y,n_z,n_w)=\dfrac{\pi^2\hbar^2}{2mL^2}(n_x^2+n_y^2+n_z^2+n_w^2)\]

since all dimensions have the same length, all dimensions have the same contribution to energy

Finding the first, second, and third lowest energy levels:

\[E_1=E(1,1,1,1)\]

Only 1 possibility, non-degenerate for E₁

\[E_2=E(2,1,1,1)=E(1,2,1,1)=E(1,1,2,1)=E(1,1,1,2)\]

4 possibilities, 4-fold degenerate for E₂

\[E_3=E(2,2,1,1)=E(2,1,2,1)=E(2,1,1,2)=E(1,2,2,1)=E(1,2,1,2)=E(1,1,2,2)\]

6 possibilities, 6-fold degenerate for E₃

Q10.22b

Compare the process of calculating the degeneracy within a 2-D box and 3-D box.

S10.22b

A particle in a 3-D box has many more degeneracies in comparison to a particle in a 2-D box. The size of a 2-D box is L x L, which gives a ground state energy of 2E₀. 2-D Box: \[E= E_0(n^{2}_x+n^{2}_y)\]

A 3-D cubic box has a size of L x L x L, and \[E=E_0(n^{2}_x+n^{2}_y+n^{2}_z)\]

Q10.22c

Calculate the degeneracies of the first four levels for a particle in a cubical box.