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Chemistry LibreTexts

1.E: The Dawn of the Quantum Theory (Exercises)

  • Page ID
    13390
  • Solutions to select questions can be found online.

    1.1A

    Sodium metal has a threshold frequency of \(4.40 × 10^{14}\) Hz. What is the kinetic energy of a photoelectron ejected from the surface of a piece of sodium when the ejecting photon is \(6.20 × 10^{14}\) Hz? What is the velocity of this photoelectron? From which region of the electromagnetic spectrum is this photon?

    1.1B

    What is the longest-wavelength electromagnetic radiation that can eject a photoelectron from silver, given that the work function is 4.73 eV? Is this in the visible range?

    Solution

    263 nm

    1.1C

    Find the longest-wavelength photon that can eject an electron from potassium, given that the work function is 2.24 eV. Is this visible electromagnetic radiation?

    1.1C

    What is the work function in eV of electrons in magnesium, if the longest-wavelength photon that can eject electrons is 337 nm?

    Solution

    3.69 eV

    1.1D

    Calculate the work function in eV of electrons in aluminum, if the longest-wavelength photon that can eject the electromagnetic is 304 nm.

    1.1E

    What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-nm electromagnetic radiation, given that the work function is 2.28 eV?

    Solution

    0.483 eV

    1.1F

    UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kinetic energy of the ejected photoelectrons?

    1.1G

    Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the work function of electrons to sodium metal?

    Solution

    2.25 eV

    1.1H

    UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV electrons. What is the work function of electrons to uranium metal?

    1.1I

    What is the wavelength of electromagnetic radiation that ejects 2.00-eV electrons from calcium metal, given that the work function is 2.71 eV? What type of electromagnetic radiation is this?

    Solution

    (a) 264 nm

    (b) Ultraviolet

    1.1J

    Find the wavelength of photons that eject 0.100-eV electrons from potassium, given that the work function is 2.24 eV. Are these photons visible?

    1.1K

    What is the maximum velocity of electrons ejected from a material by 80-nm photons, if they are bound to the material by 4.73 eV?

    Solution

    1.95 × 10 6 m/s

    1.1L

    Photoelectrons from a material with a work function of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device?

    1.1M

    A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the work function is 2.71 eV?

    Solution

    (a) \(4.02×10^{15}/s\)

    (b) 0.256 mW

    1.1N

    (a) Calculate the number of photoelectrons per second ejected from a 1.00-mm 2 area of sodium metal by 500-nm electromagnetic radiation having an intensity of \(1.30\; kW/m^2\) (the intensity of sunlight above the Earth’s atmosphere). (b) Given that the work function is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons. Where does the other power go? How can it be recovered?

    1.1O

    Red light having a wavelength of 700 nm is projected onto magnesium metal to which electrons are bound by 3.68 eV. (a) Use \(KE_ e=h\nu -\Phi\) to calculate the kinetic energy of the ejected electrons. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

    Solution

    (a) –1.90 eV–1.90 eV

    (b) Negative kinetic energy

    (c) That the electrons would be knocked free.

    Unreasonable Results

    1.1P

    (a) What is the work function of electrons to a material from which 4.00-eV electrons are ejected by 400-nm electromagnetic radiation? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

    1.2A

    1. Suppose the electron in a hydrogen atom is in the circular Bohr orbit with n = 30. How many times per second does it go around?
    2. Suppose now the electron drops to the n = 29 state, emitting a single photon. What is the frequency of this photon, in cycles per second?
    3. Comment on the relation between your answers in (a), (b) above. What would you guess the relation to be for n = 300?

    1.2B

    The \(\mu\) (muon) is a cousin of the electron, the only difference being its mass is 207 times greater. The μ has a lifetime of about 2 ms. If a beam of muons is directed at a solid, the muons will go into orbit around nuclei. The Bohr atom, with a muon replacing the electron, is a useful model for picturing this.

    1. For a nucleus of charge Ze, how large is the n = 1 μ orbit compared with the electron orbit?
    2. What is the frequency of the photon emitted by the μ in the n = 2 to n = 1 transition?
    3. For the gold nucleus, the n = 1 μ orbit is inside the nucleus. Find the frequency of the emitted photon for n = 2 to n = 1 in this case. (Hint: you’ll need the radius of the gold nucleus. Assume here that the positive charge is uniformly spread throughout the nucleus.)

    1.3

    Past Infrared region, in direction of the lower energies, the microwave region is located. In this region, radiation usually is characterized by frequency (\(\nu\)) which is expressed in units of \(\text{MHz}\), where \(\text{Hz}\) is a cycle per second. Given a microwave frequency of \(2.0 \times 10^{4}\, \text{MHz}\), calculate \(v\), \(λ\), and energy per photon for this radiation and then compare the results with figure below.

    Solution

    The frequency(v) of the microwave radiation is given and once convert to Hz get the following

    v=2.0e4MHz(1e6Hz/1MHz)= 2.0e10s-1 Hz

    now we find the wavelength using formula and get

    \[λ=\dfrac{c}{v}=\dfrac{2.998 \times 10^{8} m \,s^{-1}}{2.0 \times 10^{10} \,s^{-1} } = 1.5 \times 10^{-2}\,m \nonumber\]

    finally we use E=hv to calculate the energy

    E=hv=(6.626e-34J*s)(2.0e10s-1)= 1.3e-23J

    1.4

    Compare the Planck Distribution and the Rayleigh-Jean Distributions. For large values of \({\nu}\), which one would be greater?

    Solution

    The Planck Distribution is

    \[d{\rho}=\dfrac{8{\pi}h}{c^3}\dfrac{{\nu}^3}{e^{\dfrac{h{\nu}}{k_BT}}-1}d{\nu}\nonumber \]

    And the Rayleigh Jean Distribution is

    \[d{\rho}=\dfrac{8{\pi}{\rho^2}k_BT}{c^3}d{\nu}\nonumber \]

    For larger \({\nu}\), the Rayleigh Jean Distribution increases, while the Planck Distribution decreases because of the exponential term in the denominator outweighing the \({\nu}^3\) term.

    1.4

    Plancks principal assumption was that energies of electronic oscillator can only have values \(E=nh\nu\) and \(ΔE=h\nu\). In fact, as v→0 then ΔE→0 and E becomes continuous. It should be expected that the nonclassical Plank distribution to go over to the classical Rayleigh-Jeans distribution at low frequencies, where ΔE →0. Prove that Equation 1.2 reduces to Equation 1.1 as v→0.

    Note: The Taylor expansion of an exponential

    \[e^x \approx 1+x+\left(\dfrac{x^2}{2!}\right) + ... \nonumber \]

    can truncated to \(e^x \approx 1+x\) when \(x\) is small.

    Solution

    Important to know Planks equation and put it into use:

    dp(v,T)=Pv(T)dv=(8πh/c3)(v3dv/ehv/kbT-1)

    Note: Pv(T)dv ⇒ is the radiant energy density between frequencies v and v+dv

    Now for small x we have ex≈1+x

    and as v→0, hv/kbT→0 once we have this we get the following

    dp(v,T)=(8πh/c3)*((v3dv)/(1+(hv/kbT)-1))= 8πhv3kBTdv/c3hv = 8πv2kBTdv/c3

    and this is the classical Rayleigh-Jeans distribution.

    1.5

    The visible spectrum is in the 400-700 nm range, and contains about 40% of the sun’s radiation intensity. Using the Planck Distribution, write an integral expression that can be evaluated to give this result (do not evaluate the integral).

    Solution

    The Planck Blackbody distribution in terms of wavelength is

    \[\rho_\lambda(\lambda, T) d \lambda =\dfrac{2 hc^2}{\lambda^5}\dfrac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1} d\lambda\nonumber \]

    And so the intensity contained in the visible spectrum (from 400 nm and 700 nm) is

    \[\int^{700}_{400}\rho_\lambda(\lambda, T) d\lambda = \int^{700\, nm}_{400\, nm}\dfrac{2 hc^2}{\lambda^5}\dfrac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1} d\lambda\nonumber \]

    The intensity contained in the whole spectrum can be given by

    \[\int^{\infty}_{0}\rho_\lambda(\lambda, T) d\lambda = \int^{\infty}_{0}\dfrac{2 hc^2}{\lambda^5}\dfrac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1} d\lambda\nonumber \]

    And thus

    \[40\% =100\% \times \dfrac{\rho_{visible}}{\rho_{total}}=100 \% \times \dfrac{\int^{700}_{400}{\dfrac{8phc}{\lambda^5 \left(e^{\dfrac{hc}{\lambda k_bT}}-1\right)} d \lambda}}{\int^{\infty }_0{\dfrac{8phc}{\lambda^5\left(e^{\dfrac{hc}{\lambda k_bT}}-1\right)}d \lambda}}\nonumber \]

    1.13

    What is the frequency and energy of a single 310 nm photon?

    Solution

    Given: \(\lambda\) = 310 nm.

    To find the frequency:

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

    \[\nu=\dfrac{2.99\times10^{8} \dfrac{m}{s}}{310 \ nm}\nonumber \]

    \[\nu= 9.67\times10^{14} s^{-1} \nonumber \]

    To find the energy:

    \[E = h \nu\nonumber \]

    \[E = (6.626 \times10^{-34})\times \nu\nonumber \]

    \[E = 6.41\times 10^{-19} J\nonumber \]

    1.14

    A laser emits \(3.3\times 10^{17}\)photons per second. If the energy per photon is \(6.4\times 10^{-20}J\) per photon, find the a) wattage and b) wavelength of the laser. in what electromagnetic spectra is the laser in?

    Solution

    a)

    \(W=(6.4\times 10^{-20}J)(3.3\times 10^{17}\dfrac{1}{s})\)

    \(W=0.02112\dfrac{J}{s}\)

    b)

    \(E=h\dfrac{c}{\lambda}\)

    \(\lambda=\dfrac{hc}{E}\)

    \(\lambda=\dfrac{(3\times 10^{8}\dfrac{m}{s})(6.626\times 10^{-34}Js)}{6.4\times 10^{-20}J}\)

    \(\lambda=3.106\times 10^{-6}m\)

    c)

    infrared spectrum

    1.15

    What is the max wavelength with a given temperature of 7500K?

    Solution

    For a given temperature, the maximum wavelength allowed is given by:

    \(T = \frac{2.9\times10^{-3}\text{mK}}{\lambda_\text{max}}\)

    Given: T = 7500K

    \(7500 = \frac{2.9\times10^{-3}\text{mK}}{\lambda_\text{max}}\)

    \(\lambda_\text{max}= \frac{2.9\times10^{-3}\text{mK}}{7500 \text{K}}\)

    \(\lambda_\text{max}= 3.8\times10^{-7} \text{m}\)

    Q1.15

    A light bulb is a blackbody radiator. What temperature is required such that \(\lambda_{max} = \ 400 nm\) ?

    Solution

    \[T=\dfrac{(2.90 \times 10^{-3}\; m \cdot K)}{400 \times 10^{-9}\; m} = 7250\; K.\nonumber \]

    1.16

    An unknown elemental metal has work function of \( \Phi = 8.01 \times 10^{-19} J\). Upon illumination with UV light of wavelength 162 nm, electrons are ejected with velocity of at \(2.95 \times 10^{3} \frac {m}{s}\). What is the threshold wavelength? What is the work function in units of eV? What metal does this correspond to (you will need to consult Table B1)?

    Solution

    This question involves a bit of a trick in that neither the wavelength of radiation nor the velocity of electrons are necessary to solve for the threshold wavelength or material as requested. To solve for the threshold wavelength, we employ the concept that kinetic energy is 0 at threshold frequency and then use a relation equation to solve for threshold wavelength.

    \[ \dfrac {1}{2} mv^2 = h \nu - \Phi \tag {2-5}\nonumber \]

    So,

    \[ \nu_{threshold} = \dfrac {\Phi}{h} = 1.21 \times 10^{15} s^{-1}\nonumber \]

    and

    \[ \lambda_{threshold} = \dfrac {c}{\nu_{threshold}} = 248 nm\nonumber \]

    With a basic conversion of

    \[ 1\;J = 6.242 \times 10^{18} eV\nonumber \]

    we see that the work function is 5 eV. Using Table B1, we see that this value corresponds to Cobalt (discovered by Georg Brandt).

    1.16

    Given the work function of sodium is 1.87 eV, find the kinetic energy of the ejected electrons when light of frequency 2.3 times greater than the threshold frequency is used to excite the electrons.

    Solution

    step 1: convert work function from electron volts to joules

    \[\phi=1.87 \ eV\nonumber \]

    1 eV = \(1.602\times 10^{-19}J\)

    \(\phi\)=\(1.87 eV \times \dfrac{1.602\times10^{-19}J}{1 \ eV}\)=\(2.995 \times 10^{-19}J\)

    Step 2: Solve for the threshold frequency

    \(\phi\)=\(hf\)

    \(\dfrac{\phi}{h}=f\)

    \(\dfrac{2.995\times 10^{-19}J}{6.626\times 10^{-34}Js}=f\)

    \(4.5\times 10^{14}Hz=f\)

    Step 3: Use threshold frequency to solve for kinetic energy at desired conditions

    \(KE=h(2.3f-f)\)

    \(KE=h(1.3f)\)

    \(KE=(6.626\times 10^{-34}Js)(1.3(4.5\times 10^{14}Hz))\)

    \(KE=3.87\times 10^{-19}J\)

    Q1.17

    Find kinetic energy emitted off surface of tungsten that is radiated with radiation of 250 nm. Work function of tungsten 4.50 eV.

    Solution

    Kinetic energy is represented by

    \[E = h\nu\nonumber \]

    we then use \(c=\nu\lambda\) for the frequency to find

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

    Then substitute values to get

    \[E=\dfrac{(6.626 \times 10^{-34}\; J \cdot s)(3.00 \times 10^8 m/s)}{250 \times 10^{-9}\; m} = 7.95 \times 10^{-19}\; J\nonumber \]

    Convert to eV, to get

    \[E=(7.95 \times 10^{-19}\; J) \dfrac{1 \;eV}{1.6 \times 10^{-19}\;J} = 4.97\; eV.\nonumber \]

    Use \(KE_e=h\nu–\Phi\) to finally get

    \[KE_e=h\nu – \Phi = 4.97 \;eV – 4.50 \;eV = 0.47\; eV.\nonumber \]

    1.18

    A smooth silver Thanksgiving platter and serving spoon (the pilgrims had knives and spoons but no forks) are irradiated with light of wavelength 317 nm. The work function is \(\Phi = 6.825 \times 10^{-19} J\) . What is the kinetic energy of the ejected electrons [eV]? The threshold frequency?

    Solution

    We first solve for the threshold frequency.

    \[ \dfrac {hc}{\lambda} = \Phi \ = h \nu\nonumber \]

    Rearrange to solve for \( \nu\)

    \[\begin{align*} \nu &= \dfrac {\Phi}{h} \\[4pt] &= 1.03 \times 10^{15} s^{-1} \end{align*}\nonumber \]

    Now we solve for kinetic energy.

    \[ \dfrac {1}{2} mv^2 = h \nu - \Phi \tag {2-5}\nonumber \]

    where \[ \nu = \nu_{radiation}\nonumber \] and we recall that

    \[ \nu = \dfrac {c}{\lambda_{radiation}}\nonumber \]

    Using the right hand side of that kinetic energy equation, we find the result to be

    \[ {KE} = 2.55 \times 10^{-19}\nonumber \]

    Q1.18

    When a clean surface of silver is irradiated with light of wavelength 255 nm, the work function of ejected electrons is 4.18 eV. Calculate the kinetic energy in eV of the silver and the threshold frequency.

    Solution

    Kinetic Energy of the electrons can be represented with the formula

    \(KE= h\nu - \Phi \)

    We have to solve for the Kinetic energy in eV

    \(KE= h\nu - \Phi \)

    substituting known values gives

    \[ KE= (6.626\times 10^-34 Js )(\dfrac{3\times 10^8 m/s}{255\times 10^-9 m})- 6.69* 10^-19 J\nonumber \]

    \[ KE = 1.105 \times 10^-19 J \approx .690 eV\nonumber \]

    The second part of the question asks us to solve for the threshold frequency

    \( \nu_o= \dfrac{\Phi}{h}\)

    \( \nu_o= \dfrac{6.69\times 10^-19 J}{6.626\times 10^-34 J*s}\)

    \( = 1.01\times 10^{15} Hz\)

    Q1.21

    A line in the Paschen series of hydrogen has a wavelength of 1.01 x 10-6 m. Find the original energy of the electron.

    Solution

    For the Paschen series n1= 3. To find n2 we have to use the Rhydberg formula:

    \[\dfrac{1}{\lambda} = R_H \times \left(\dfrac{1}{n_1^2} + \dfrac{1}{n_2^2}\right)\nonumber \]

    substituting our known values will give us

    \[\dfrac{1}{1.01\times 10^{-6} m } = 109677 cm^{-1} \left(\dfrac{1}{3^2} + \dfrac{1}{n_2^2}\right)\nonumber \]

    converting our units and using algebra gives us

    \(0.0903 = \left(\dfrac{1}{9} - \dfrac{1}{n_2^2}\right)\)

    where

    \(n_2 = 6.93 \approx 7 \)

    We approximate to 7 since \(n\) is an integer.

    1.22

    How does the energy change when a particle absorbs and releases a photon? Show the effects on the state that the particle is in and the energy itself.

    Solution

    The energy would increase when a photon is absorbed and decrease when a photon is released. We have two equations

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

    \[\dfrac{1}{\lambda} = 109680\left(\dfrac{1}{n_1} - \dfrac{1}{n_2}\right)\nonumber \]

    When a photon is absorbed, \(\lambda\) is positive and when a photon is released \(\lambda\) is negative. From the first equation, we can see that \(E\) only depends on the sign of \(\lambda\). So when a photon is absorbed, energy is positive (increases) and when it is released, energy is negative (decreases).

    The second equation shows that when \(\lambda\) is negative, \(n_1\) must be greater than \(n_2\) so the final state is at a lower quantum number than the initial and vice versa for when a photon is absorbed.

    1.23

    Show that the (a) wavelength of 100 nm occurs within the Lyman series, that (b) wavelength of 500 nm occurs within the Balmer series, and that (c) wavelength of 1000 nm occurs within the Paschen series. Identify the spectral regions to which these wavelengths correspond.

    Solution

    We can show the where the wavelengths occurs by calculate the maximum and minimum wavelengths of each series using the Rydberg formula.

    a) Lyman Series:

    \[Max: \dfrac{1}{\lambda} = 109680\left(1 - \dfrac{1}{2^2}\right)cm^{-1}\nonumber \]

    \[\lambda = 121.6 nm\nonumber \]

    \[Min: \dfrac{1}{\lambda} = 109680\left(1 - \dfrac{1}{\infty}\right)cm^{-1}\nonumber \]

    \[\lambda = 91.2 nm\nonumber \]

    The range for the Lyman series from 91.2 nm to 121.6 nm, therefore a wavelength of 100 nm occurs within the Lyman series. This corresponds to the ultraviolent region of the spectrum.

    b) Balmer Series:

    \[Max: \dfrac{1}{\lambda} = 109680\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)cm^{-1}\nonumber \]

    \[\lambda = 656.5 nm\nonumber \]

    \[Min: \dfrac{1}{\lambda} = 109680\left(\dfrac{1}{2^2} - \dfrac{1}{\infty}\right)cm^{-1}\nonumber \]

    \[\lambda = 364.7 nm\nonumber \]

    The range for the Balmer series from 364.7 nm to 656.5 nm, therefore a wavelength of 500 nm occurs within the Balmer series. This corresponds to the near ultraviolent region of the spectrum.

    c) Paschen Series:

    \[Max: \dfrac{1}{\lambda} = 109680\left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right)cm^{-1}\nonumber \]

    \[\lambda = 1875.6 nm\nonumber \]

    \[Min: \dfrac{1}{\lambda} = 109680\left(\dfrac{1}{3^2} - \dfrac{1}{\infty}\right)cm^{-1}\nonumber \]

    \[\lambda = 820.6 nm\nonumber \]

    The range for the Paschen series from 820.6 nm to 1875.6 nm, therefore a wavelength of 1000 nm occurs within the Paschen series. This corresponds to the near infrared region of the spectrum.

    1.24

    Calculate the wavelength and the energy of a photon associated with the series limit of the Balmer series.

    Solution

    First find the minimum wavelength for the Balmer series.

    \[\dfrac{1}{\lambda} \ = 109,680cm^{-1}\Big(\dfrac{1}{2^2} \ - \ \dfrac{1}{\infty}\Big)\nonumber \]

    \[\lambda \ = \ 364.7 nm\nonumber \]

    Now we can use the wavelength to find the energy.

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

    \[E = \dfrac{(6.626 \ \times \ 10^{-34})(3 \ \times \ 10^8)}{364.7 \ \times \ 10^{-9}}\nonumber \]

    \[E = 5.45 \ \times \ 10^{-19}J\nonumber \]

    1.25

    For the following particles (a) an electron with a kinetic energy of 50eV, (b) a proton with a kinetic energy of 50eV , and (c) an electron in the second Bohr orbit of a hydrogen atom, calculate the de Broglie wavelength of each.

    Solution

    We use \(\lambda \ = \dfrac{h}{p}\) in all cases to find \(\lambda\).

    a. \[KE \ = \dfrac{mv^2}{2}\nonumber \]

    \[50eV\left(\dfrac{1.602\times10^{-19}J}{1 eV}\right) \ = \ \dfrac{(v^2)(9.109\times10^{-31}kg)}{2}\nonumber \]

    \[v = 4.19\times10^6m\centerdot s^{-1}\nonumber \]

    So

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

    \[\ = \ \dfrac{6.626\times10^{-34}J\centerdot s}{(9.109\times10^{-31}kg)(4.19\times10^6m\centerdot s^{-1})}\nonumber \]

    \[\ = 1.23\times10^{-10}m = 0.123nm\nonumber \]

    b. Replace \(m_e\) with \(m_p\) in (a) to find \(\lambda = 2.86\times10^{-3} nm\).

    c. We must first determine the velocity of an electron in the second Bohr orbit of a hydrogen atom. The velocity of an electron is given by the following equation:

    \[v = \dfrac{nh}{2(\pi)m_{e}r}\nonumber \]

    and we know

    \[r = \dfrac{\epsilon_0h^2n^2}{(\pi)m_ee^2}\nonumber \]

    substituting the two equations we find that

    \[v \ = \ \dfrac{e^2}{2nh\epsilon_0}\nonumber \]

    For n = 2, because we are talking about the second orbit

    \[v \ = \ \dfrac{(1.602\times10^{-19}C)^2}{2(2)(6.626\times10^{-34}J\centerdot s)(8.854\times10^{-12}C^2J^{-1}m^{-1})}\nonumber \]

    \[\ = \ 1.09\times10^6m\centerdot s^{-1}\nonumber \]

    So

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

    \[\ = \ \dfrac{6.626\times10^{-34}J\centerdot s}{(9.109\times10^{-31}kg)(1.09\times10^6m\centerdot s^{-1}}\nonumber \]

    \[\ = \ 6.64\times10^{-10}m = 0.664nm\nonumber \]

    Q1.26

    1. What is the velocity and wavelength of an electron with a voltage increase of 75 V?
    2. What is the momentum of an electron with a de Broglie wavelength of 20 nm?(mass of an electron is 9.109x10-31kg)
    Solution

    \[a) (electron charge)x(potential) = KE\nonumber \]

    \[(1.602x10^{-19}C)(75V) = KE\nonumber \]

    \[KE = 1.2x10^{-17}J\nonumber \]

    \[KE = (1/2)mv^2\nonumber \]

    \[v = \sqrt{\dfrac{2(KE)}{m}} = \sqrt{\dfrac{2(1.2x10^{-17}J)}{(9.109x10^{-31}kg)}} = 5.133x10^{6}m*s^{-1}\nonumber \]

    \[\lambda = \dfrac{h}{mv} = \dfrac{6.626x10^{-34}J*S}{(9.109x10^{-31}kg)(5.133x10^{6}m*s^{-1})} = 1.2267 m\nonumber \]

    \[b) \lambda = \dfrac{h}{p}\nonumber \]

    \[p= \dfrac{h}{\lambda} = 6.626x10^{-34}J*S/ 20x10^{-6}m = 3.313x10^{-29}kg*m*s^{-1}\nonumber \]

    1.27

    Through what potential a proton must initially at rest fall so its de Broglie wavelength is 1.83×10-10 m ?

    Q1.28

    Calculate the energy and wavelength associated with a \(\beta\) particle that has fallen through a potential difference of 3.2V. Take the mass of a \(\beta\) particle to be \(9.1\times10^{-31}\text{kg}\).

    Solution

    A beta particle is an electron, so it has a -1 charge.

    \[KE = \beta\;particle charge\;x\;Potential = \vert-1.602\times10^{-19}\text{C}\vert\;x\; 3.2\text{V}\nonumber \]

    \[KE = 5.126\times10^{-19}\text{J per }\beta\text{ particle}\nonumber \]

     

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

    \[KE = \dfrac{mv^2}{2} = \dfrac{p^2}{2m}\nonumber \]

    \[p = \sqrt{2\;KE\;m} = \sqrt{2\;x\;5.126\times10^{-19}\text{J}\;x\;9.1\times10^{-31}\text{kg}}\nonumber \]

    \[p = 5.66\times10^{-25}\text{kg m }s^{-1}\nonumber \]

    \[\lambda = \dfrac{6.626\times10^{-34}/text{J s}}{5.66\times10^{-25}\text{kg m s^-1}} = 6.86\times10^{-10}\text{m}\nonumber \]

    Q1.28

    If a proton is going through a potential difference of 3.0V, what is the momentum and wavelength associated with this proton? (mass of a proton is equal to 1.6726x10-27kg)

    Solution

    \[(charge)*(potential) = KE\nonumber \]

    \[charge = 1.602x10^{-19}C\nonumber \]

    \[(1.602x10^{-19}C)*(3.0V) = KE\nonumber \]

    \[KE = 4.806x10^{-19}J\nonumber \]

    \[KE = \dfrac{p^2}{2m} \nonumber \]

    \[p= \sqrt{2(KE)m} = \sqrt{2(4.806x10^{-19}J)(1.6726x10^{-27}kg)} = 4.01x10^{-23}kg*m*s^{-1}\nonumber \]

    \[\lambda = h/p = \dfrac{6.626x10^{-34}J*S}{4.01x10^{-23}kg*m*s^{-1}} = 1.65x10^{-11}m = 0.165pm\nonumber \]

    1.29

    Neutron diffraction is a modern technique to study structure. In the neutron diffraction, a collimated beam of neutrons was generated at some temperature from high-energy neutron source. This is achieved at several accelerator facilities around the world. If the speed of neutron is vn = (3kBT/m)1/2, m is the mass of neutron. What is the required temperature so that neutrons have a de Broglie wavelength of 200 pm? The mass of a neutron to be 1.67×10-27 kg

    1.29

    While studying quantum mechanics one day, you wondered what temperature would be required for the Jumbo Jawbreaker you were about to eat to have a de Broglie wavelength of \(1.9 \times 10^{-24}\) meters? Assuming that the speed of a Jumbo Jawbreaker can be calculated from the equation \(\nu_n = (\dfrac{3k_BT}{m})^{\dfrac{1}{2}}\). You quickly measure the mass of your Jumbo Jawbreaker and found it to be \(0.1kg\).

    Solution

    Knowing that the de Broglie wavelength has the form,

    \[ \lambda = \dfrac{h}{m\nu_n}\nonumber \]

    we can substitute the given equation for speed into the de Broglie wavelength equation

    \[ \lambda = \dfrac{h}{(3mk_BT)^{\dfrac{1}{2}}} \nonumber \]

    rearrange to solve for temperature

    \[ T = \dfrac{h^2}{3mk_B\lambda^2}\nonumber \]

    Substituting in constants we can solve for temperature in Kelvin. Using \(h = 6.626 \times 10^{-34} J*s\), \(m = 0.1kg\), \(k_B = 1.381 \times 10^{-21} J*K^{-1}\), and \(\lambda = 1.9 \times 10^{-24}\) meters.

    We find that

    \[ T = \dfrac{(6.626 \times 10^{-34} J*s)^2}{3(0.1kg)(1.381 \times 10^{-21} J*K^{-1})(1.9 \times 10^{-24})^2}\nonumber \]

    Therefore

    \[T = 293.5K \\ T=20.35 \ ^\circ C\nonumber \]

    1.30

    For linear motion, show that a small change in the momentum, \(\Delta\text{p}\), changes a change in kinetic energy, \(\Delta\text{KE}\), of

    \[\Delta\text{KE} = \dfrac{p_{0}}{m}\Delta\text{p}\nonumber \]

    where \(p_{0}\) is initial momentum.

    Solution

    Since \(\Delta\text{p} = dp\) and \(\Delta\text{KE} = dKE\),

    \[KE = \dfrac{p^2}{2m}\nonumber \]

    \[dKE = \dfrac{p_{0}}{m}dp\nonumber \]

    \[\Delta\text{KE} = \dfrac{p_{0}}{m}\Delta\text{p}\nonumber \]

    1.31

    Derive the Bohr formula for \(\dfrac{1}{\lambda_{vac}}\) for a multi-proton and single electron atom such as \(\ce{He^{+}}\) or \(\ce{Li^{2+}}\).

    Solution

    The number of protons (Z) in the nucleus interact with the single electron with the same coulomb force \((f)\). The total force of a nucleus with charge Z can be written as the sum of each proton individually interacting with the electron.

    \[ f_{Total} = \displaystyle\sum\limits_{i=0}^Z \dfrac{e^2}{4r^2\pi\epsilon_\circ}\nonumber \]

    Simplifying this expression we find that

    \[f_{Total} = \dfrac{Ze^2}{4r^2\pi\epsilon_\circ}\nonumber \]

    To prevent the electron from spiraling into or away from the nucleus, the centrifugal force \(f = \dfrac{m_ev^2}{r}\) is equal to the Coulombic force. Therefore

    \[ \dfrac{Ze^2}{4r^2\pi\epsilon_\circ}=\dfrac{m_e\nu^2}{r}\nonumber \]

    For stability purposes a condition requires electrons to have a set number of complete wavelengths around the circumference of the orbit or

    \[2\pi r = n\lambda \ , \ where \ n = 1,2,3...\nonumber \]

    using the de Broglie wavelength formula \(\lambda = \dfrac{h}{p} = \dfrac{h}{m\nu}\) we find that \[m_e\nu r = \dfrac{nh}{2\pi}\nonumber \] Solving for \(\nu\) and substituting into our force relationship \( \dfrac{Ze^2}{4r^2\pi\epsilon_\circ}=\dfrac{m_e\nu^2}{r}\) We find that

    \[r = \dfrac{n^2h^2\epsilon_\circ}{m_ee^2Z\pi}\nonumber \]

    Now solving for the total energy of the system

    \[E = KE +V(r) \\ = \dfrac{1}{2}m_e\nu^2 - \dfrac{Ze^2}{4r\pi\epsilon_\circ}\nonumber \]

    Substituting in \(m_e\nu^2\) found above into the kinetic energy portion we find

    \[ E = \dfrac{Ze^2}{8r\pi\epsilon_\circ} - \dfrac{Ze^2}{4r\pi\epsilon_\circ} = -\dfrac{Ze^2}{8r\pi\epsilon_\circ}\nonumber \]

    Substituting \(r\) from above we quantize the energy such that

    \[E_n = \dfrac{-Z^2m_ee^4}{8n^2h^2\epsilon_\circ^2}\nonumber \]

    Since this energy is quantized, the change in energy states will occur where electrons are excited by light or \(h\nu\) into higher quantum states. Therefore

    \[\Delta E = \dfrac{-Z^2m_ee^4}{8h^2\epsilon_\circ^2} \big(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\big)=h\nu\nonumber \]

    Finally solve for \(\dfrac{1}{\lambda_{vac}}\) remembering that \(h\nu = \dfrac{hc}{\lambda_{vac}}\) where \(c\) is the speed of light. We obtain our final solution

    \[\boxed{\dfrac{1}{\lambda_{vac}} = \dfrac{-Z^2m_ee^4}{8h^3c\epsilon_\circ^2} \big(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\big)}\nonumber \]

    1.32

    The series in the He+ spectrum that corresponds to the set of transitions where the electron falls from a higher level into the n = 4 state is called the Pickering series, an important series in solar astronomy. Derive the formula for the wavelengths of the observed lines in this series. In what region of the spectrum does it occur?

    Solution

    If we derive the Bohr formula for \(\tilde{v} = Z^2R_H(\dfrac{1}{(n^2)_1} - \dfrac{1}{(n^2)_2}\)

    In the Pickering series the helium spectrum is in Z = 2 and \(n_2\) = 4

    \[\tilde{v} = 4(109,680 cm^-1)((\dfrac{1}{4^2} - \dfrac{1}{n^2_1}, where n_1 = 5, 6, 7, 8....\nonumber \]

    \[ n_1 = 5, \tilde{v}\ = 9871 cm^-1 \nonumber \]

    or

    \[\lambda = 1.013 * 10^{-6} meters \nonumber \]

    1.33A

    Using the Bohr model, find the third ionization energy for the Lithium atom in eV and in J.

    Solution

    Energy transitions for a hydrogen like atom are given by ΔE = Z2Ry(1/ni2 - 1/nf2 )

    where Z is the atomic number and Ry is 13.6 eV

    When a hydrogen like atom is ionized, the electron transitions to its highest bound state, at n = infinity,so its quantum number nf goes to infinity, making 1/nf2 = 0.

    So Eionization = (3)2(13.6)(1/(1)2 - 0) = 122.9 eV.

    122.9 eV * 1.6 *10-19 = 1.96*10-17 J

    1.33B

    Find the ionization energy in eV and \(kJ \centerdot mol^{-1}\) of singly ionized helium in the n=3 state, using Bohr theory

    Solution

    To find the ionization energy of helium, consider the case where we move an electron from the n=3 state to an infinite distance from the nucleus.

    Using the Bohr formula for \(\tilde{v}\).

    \[\tilde{v}=Z^2R_H(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\nonumber \]

    \[\tilde{v}=2^2(109680 cm^{-1})(\dfrac{1}{3^2}-\dfrac{1}{\infty^2})\nonumber \]

    \[\tilde{v}=4.87467 \times 10^4 cm^{-1}\nonumber \]

    Then plugging into E=hc\(\tilde{v}\)

    \[E= (6.626 \times 10^{-34} J \centerdot s)(2.998 \times 10^8 m \centerdot s^{-1})(4.87467 \times 10^6 m^{-1})\nonumber \]

    \[E=9.68 \times 10^{-19} J = 583 kJ \centerdot mol^{-1} = 6 eV\nonumber \]

    1.34A

    The speed of electron in an nth Bohr orbit is given by the equation:

    \[v = \dfrac{e^2}{2\epsilon_0nh}\nonumber \]

    The force acting between an electron and proton a distance r from one another is given by Coulomb's law:

    \[f = \dfrac{e^2}{4\pi\epsilon_0{r^2}}\nonumber \]

    The centrifugal force acts in opposition to the Coulombic force and is given by the equation:

    \[f = \dfrac{mv^2}{r}\nonumber \]

    Find the values of \(v\) for the Bohr orbits of n = 4, n = 5, and n = 6, and find the total force in an atom between a proton and electron a distance of 5 x 10-11 m away from one another, with the electron moving at a speed of 2 x 106 m/s.

    Solution

    To find \(v\) simply substitute the values for n into the equation:

    \[v = \dfrac{e^2}{2\epsilon_0nh}\nonumber \]

    For the values of \(v\) at n = 4, n = 5, n = 6, we get

    n = 1

    \(v_1= 546,923 m*s{^-1}\)

    n = 2

    \(v_2 = 437,538 m*s{^-1}\)

    n = 3

    \(v_3 = 364,615 m*s{^-1}\)

    To find the force between a proton and electron, simply subtract the Coulombic force from the Centrifugal force and substitute appropriate values for the constants:

    \[f = \dfrac{mv^2}{r} - \dfrac{e^2}{4\pi\epsilon_0{r^2}}\nonumber \]

    For which we attain:

    \[f = 7.2875 x 10^{-8} N\nonumber \]

    1.34B

    Prove that the speed of electron in an nth Bohr orbit is \(v = \dfrac{e^2}{2\epsilon_0nh}\)

    Then find the first few values of \(v\) the Bohr orbit.

    Solution

    First we have to know that the angular moment of the electron revolving in the \(n\)th Bohr orbit is quantized then

    \(mvr = \dfrac{nh}{2\pi}\), where r = radius of the \(n\)th Bohr orbit

    Kinetic energy of the electron is given as \(\dfrac{mv^2}{2} = \dfrac{e^2}{2(4\pi\epsilon_0)r}\)

    So the radius, r must equal \(r = \dfrac{e^2}{(4\pi\epsilon_0)mv^2}\)

    Now after substituting the value above into the first equation, we get \(mv(\dfrac{e^2}{(4\pi\epsilon_0)mv^2} = \dfrac{nh}{2\pi}\)

    Thus the speed of the electron in the \(n\) Bohr orbit is \(v = \dfrac{e^2}{2\epsilon_0nh}\)

    For the first few values of \(v\) in the \(n\)th Bohr orbit, we get

    n = 1

    \(v= 2.188 * 10^{6} m*s{^-1}\)

    n = 2

    \(v_2 = 1.094*10^{6} m*s{^-1}\)

    n = 3

    \(v_3 = 7.292 * 10^{5} m*s{^-1}\)

    1.35

    What is the uncertainty in an electron's position if the uncertainty in measuring its velocity is 5 \(m \centerdot s^{-1}\).

    Solution

    According to the Heisenburg Uncertainty Principle

    \[\Delta x \Delta p \geq \dfrac{\hbar}{2}\nonumber \]

    \[\Delta x \geq \dfrac{\hbar}{2\Delta p}\nonumber \]

    Then by definition \(\Delta p = m \Delta v\)

    \[\Delta x \geq \dfrac{\hbar}{2m \Delta v}\nonumber \]

    \[\Delta x \geq \dfrac{6.626 \times 10^{-34} J \centerdot s}{4\pi (9.109 \times 10^{-31} kg)(5 m \centerdot s^{-1})}\nonumber \]

    \[\Delta x \geq 1.16 \times 10^{-5}m\nonumber \]

    1.35

    What is the uncertainty in the speed of an electron if we locate it to within 50 pm?

    Solution

    It is known that the uncertainty of momentum is given by the expression $$\Delta{p} = m\Delta{v}$$and Heisenberg’s Uncertainty Principle states that $$\Delta{x}\Delta{p} \ge h$$Then

    $$\Delta{x}(m\Delta{v}) \ge h$$

    $$\Delta{v} \ge \frac{h}{m\Delta{x}}$$

    $$\Delta{v} \ge \frac{6.626 \cdot 10^{-34} J \cdot s}{(9.109 \cdot 10^{-31} kg)(50 \cdot 10^{-12} m)}$$

    $$\Delta{v} \ge 1.45 \cdot 10^7 m \cdot s$$

    1.35

    Using the Heisenberg Uncertainty Principle, what is the uncertainty in position of an electron moving at the speed of light.

    Solution

    \[\Delta x \Delta p \ge \dfrac{h}{4\pi} \nonumber \]

    From classical Physics and the problem we know

    \[\Delta p=m \cdot \Delta v\nonumber \]

    \[ m_e = 9.109 x 10^{-11} kg\nonumber \]

    \[and\nonumber \]

    \[\Delta v = 3 x 10^{-8} \dfrac{m}{s}\nonumber \]

    through substitution we find:

    \[\Delta x \cdot 9.109 x 10^{-11}kg \cdot 3x10^{-8} \dfrac{m}{s} \ge 6.626 x 10^{-34} \dfrac{kg\cdot m^2}{s}\nonumber \]

    \[ \Delta x \cdot 2.733 x 10^{-18} \dfrac{kg \cdot m}{s} \ge 5.204 x 10^{-34} \dfrac{kg \cdot m^2}{s}\nonumber \]

    \[\Delta x \ge 1.904 x 10^{-16 }m\nonumber \]

    1.35

    If we know the velocity of an electron to within \(3.5 \times 10^{7}\dfrac{m}{s}\), then what is the uncertainty in its position?

    Solution

    Using the Heisenberg Uncertainty Principle,

    \[\Delta x \Delta p \geq h = \Delta x \times m\Delta v \geq h\nonumber \]

    and rearranging to solve for uncertainty in velocity,

    \[\Delta x \geq \dfrac{h}{m\Delta v}\nonumber \]

    we can use \(h = 6.626 \times 10^{-34} J \cdot s\), \(m = 9.109 \times 10^{-31} kg\), and \(\Delta v = 3.5 \times 10^{7}\dfrac{m}{s}\) and find that

    \[ \Delta x \geq \dfrac{(6.626 \times 10^{-34} J \cdot s)}{(9.109 \times 10^{-31} kg)(3.5 \times 10^{7}\dfrac{m}{s})}\nonumber \]

    and thus

    \[\Delta x \geq 2.078 \times 10^{-11} meters\nonumber \]

    1.35

    If a proton is located to within 1 angstrom, what is its uncertainty in velocity?

    Solution

    The Heisenberg uncertainty principle states

    ΔxΔp = h/4π

    ΔxmΔv = h/4π

    Δv = h/(4mpπΔx) where mp is the mass of a proton

    x ~ Δx the uncertainty in position is on the same order as the location it is confined to, here 1 angstrom

    Δv = h/(4mpπx) = (6.626*10-34)/(4*(1.67*10-27)*3.14*10-10)

    = 315.7 m/s

    1.36

    If the position of an electron is within the 10pm interval, what is the uncertainty of the momentum? Is this value similar to that of an electron in the first Bohr orbit?

    Solution

    According to the uncertainty principle for position and momentum,

    ∆\(x\)∆\(p\)≥\(h\)

    \[\Delta p≥ \dfrac{h}{∆x}\nonumber \]

    by substituting the respective values we get,

    \[\Delta p≥ \dfrac{6.626 \times 10^{-34} J.s}{10.0 \times 10^{-12} m}\nonumber \]

    ≥\(2.9 \times 10^{-23} kg.m.s^-1\)

    Therefore, uncertainty in the momentum of an electron will be \(2.9 \times 10^{-23} kg.m.s^-1\).

    We can calculate the momentum of an electron in the first Bohr radius by using \(v\) since we know that,

    \(p\) = \(m\)\(e\)\(v\)

    = (\(9.109 \times 10^{-31} kg\))(\(2.188 \times 10^{6} m.s^-1\))

    =(\(1.992 \times 10^{-24} kg.m.s^-1\))

    The uncertainty of the momentum of an electron somewhere in a 10pm interval is greater that the momentum of an electron in the first Bohr radius.

    1.37

    The uncertainty principle applies to position and momentum:

    \[\Delta x \Delta p \geq h\nonumber \]

    Show that both sides of this expression have the same units.

    Solution

    The units for position are \(m\) and the units for momentum are \(N\cdot s\). The units for Planck's Constant are \(J\cdot s\). A \(J\) is equal to a \(N\cdot m\) so the units are the same.

    1.37

    It is known that the Heisenberg’s Uncertainty Principle is given by the expression:

    $$\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}$$

    Show that the left side has the same unit as the right side for this expression.

    Solution

    The unit of momentum p is kg*m/s, the unit of x is m, so the unit of the product of p and x is kg*m2/s.

    The Planck's constant has unit of J*s.

    Since J = kg*m2/s2, J*s = kg*m2/s, therefore, the unit for both sides of this expression is the same.

    1.37

    The Heisenberg Uncertainty Principle

    \[\Delta x\Delta p \ge \dfrac{h}{4 \pi}\nonumber \]

    Show both sides have the same units

    Solution

    \[\Delta x = meters\nonumber \]

    \[\Delta p = m \Delta v\nonumber \]

    Planck's Constant has units of \[J \cdot s\nonumber \]

    and

    \[J= \dfrac{kg \cdot m^2}{s^2}\nonumber \]

    so apply these equations we get

    \[m \dfrac{kg\cdot m^2}{s} \ge \dfrac{kg \cdot m^2\cdot s}{s^2}\nonumber \]

    and simplify

    \[\dfrac{kg \cdot m^2}{s} \ge \dfrac{kg \cdot m^2}{s}\nonumber \]

    Therefore Both sides have same units

    1.38

    The relationship between energy and time can be seen through the following uncertainty principle: ∆E∆\(t\) ≥ \(h\). Through this relationship, it can be interpreted that a particle of mass \(m\) (\(E\)=\(m\)c2) can come from nothing and return to nothing within a time ∆t ≤ \(h\)/(\(m\)c2). A real particle is one that lasts for time (\(\Delta t\)) or more; likewise, a particle that lasts for less than time (\(\Delta t\)) are called virtual particles. For a charged subatomic particle, a pion, the mass is \(2.5 \times 10^{-28} kg\). For a pion to be considered a real particle, what is its minimum lifetime?

    Solution

    Based on the uncertainty principle for energy and time:

    ∆E∆\(t\) ≥ \(h\).

    \[\Delta t≥ \dfrac{h}{mc^2}\nonumber \]

    therefore \(E\)=\(m\)c2 . By plugging in the values, you get

    \[\Delta t≥ \dfrac{6.626 \times 10^{-34} Js}{(2.5 \times 10^{-28} kg)(2.998 \times 10^{8} ms^-1) ^2}\nonumber \]

    ≥\(2.9 \times 10^{-23} s\)

    Therefore, the minimum lifetime if the pion is to be considered a real particle will be \(2.9 \times 10^{-23} s\).