Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Modern Physics Solutions

1. It requires 13.6 eV to ionize a hydrogen atom. What wavelength of light would we need to do this? (1 eV = 1.602 × 10^-19 J)

   The energy of a particular wavelength of light is given by
   

   E = hf = hc/λ .

   Now hc = 1241 nm/eV in convenient units, so
   

   λ = hc / E = (1241 nm/eV) / (13.6 eV) = 91 nm .

   

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2. In order to break a chemical bond in the molecules of human skin, causing sunburn, a photon of energy of about 3.5 eV is required. To what wavelength does this correspond?

   The energy of a particular wavelength of light is given by
   

   E = hf = hc/λ .

   Now hc = 1241 nm/eV in convenient units, so
   

   λ = hc / E = (1241 nm/eV) / (3.5 eV) = 355 nm .

   This is in the ultraviolet range which is why weather reports mentions the UV index.
   

   

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3. Electrons with maximum KE of 3.0 eV are ejected from a metal surface by ultraviolet radiation of wavelength 150 nm. Determine the work function and the corresponding threshold wavelength of the metal.
   

   The photoelectric effect is governed by the equation

   K = hf - φ = hc/λ - φ,             (1)

   where K is the kinetic energy of the electron, φ is the work function of the particular metal, and hc = 1241 nm/eV in convenient units. Solving for φ, we get
   

   φ = hc/λ - K = (1241 nm/eV) / 150 nm - 3.0 eV = 5.3 eV .

   The threshold wavelength is the wavelength for which the photoelectrons have no kinetic energy, thus from equation (1) we find
   

   λ_Threshold = hc / φ = (1241 nm/eV ) / 5.3 eV = 235 nm .

   

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4. An electron is accelerated through a potential difference of 20000 Volts. It then collides with a metal target. What is the minimum wavelength that an emitted photon can have?

   The energy of the electron will be 20,000 eV. If all the electron's energy is converted to light energy in the form of one photon, the wavelength of the photon is given by
   

   λ = hc/E = (1241 nm/eV) / (20,000 eV) = 0.062 nm .

   Notice that if less energy goes to the photon, the photon's wavelength will be longer. Thus 0.062 nm is the minimum wavelength possible for the photon.
   

   

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5. What wavelength must electromagnetic radiation have if the photon is to have the same momentum as an electron moving with speed 2.0 × 10⁵ m/s. The mass of an electron is 9.11 × 10^-31 kg.

   The momentum of a particle with mass is given by p = mv; the momentum of the massless photon is p = h/λ. Equating the two we find
   

   λ = h/mv = (6.63 × 10^-34 J-s) / (9.11 × 10^-31 kg)(2.0 × 10⁵ m/s) = 3.6 nm .

   

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6. What is the maximum energy that a free electron (initially stationary) can acquire in a collision with a photon of energy 4.0 × 10³ eV?

   Collisions between photons and electrons are called Compton scattering. In the collision, the energy and thus the wavelength of the scattered photon changes by an amount governed by Compton's formula
   

   Dl = (h/m_ec)[1 - cos(θ)] ,

   where θ is the scattering angle. The biggest change in the photon's wavelength and hence in the energy it gives to the electron occurs when θ = 180°. In this case Dl = 2h/m_ec. This occurs when the photon is scattered directly backwards. If λ₀ is the original wavelength of the photon, the change in the photon's energy is
   

   ΔE = E₀ - E_f = hc/λ₀ - hc/(λ₀ + Dl) .

   If we assume that Dl is small compared to λ₀, we can expand the above equation as
   

   E	>> hc/λ0 - (hc/λ0)(1 - Dl)
   	= (hc/λ02) Dl
   	= (hc/λ0)2 (Dl/hc)
   	= E02(Dl/hc)
   	= E02(2/mec2)
   	= (4000 eV)2(2)/(9.11 × 10-31 kg)(2.998 × 108 m/s)2
   	= 62.6 eV

   

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7. A photon of initial wavelength 0.0400 nm suffers two successive collisions with two electrons. The deflection in the first collision is 90° and in the second collision is 60°. What is the is the final wavelength of the photon?
   

   Collisions between photons and electrons are called Compton scattering. In the collision, the energy and thus the wavelength of the scattered photon changes by an amount governed by Compton's formula
   

   Dl = (h/m_ec)[1 - cos(θ)] ,

   where θ is the scattering angle.

   In the first collision,
   

   λ₁ = (h/m_ec)[1 - cos(90°)] = h/m_ec = 2.43 × 10^-3 nm .

   In the second collision,
   

   λ₂ = (h/m_ec)[1 - cos(60°)] = ½h/m_ec = 1.22 × 10^-3 nm .

   So the new wavelength of the photon is
   

   λ_new = λ_old + λ₁ + λ₂ = 0.0400 nm + 0.0024 nm + 0.0012 nm = 0.0436 nm .

   

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8. What must be the energy of an electron if its wavelength is to equal the wavelength of visible light of about 550 nm? (E = p²/2m , m_e = 9.11 × 10^-31 kg)
   

   According to the de Broglie formula, the wavelength of a massive particle is given by
   

   λ = h/p = h/m_ev .           (1)

   The kinetic energy is related to v by E = ½m_ev². Using equation (1) we eliminate v in favour of λ and we find

   E	= ½(h2 / meλ2)
   	= ½(6.626 × 10-34 J-s)2 / (9.11 × 10-11 kg)(550 × 10-9 m)2
   	= 7.97 × 10-25 J
   	= 4.97 × 10-6 eV

   

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9. A photon and an electron each have an energy of 6.0 × 10³ eV. What are their wavelengths?

   The wavelength of the photon is given by
   

   λ = hc/E = (1241 nm/eV) / (6000 eV) = 0.207 eV .

   According to the de Broglie formula, the wavelength of a massive particle is given by
   

   λ_e = h/p = h/m_ev .            (1)

   The kinetic energy is related to v by E = ½m_ev², or v = [2E/m_e]^½. Substituting this into equation (1), we find
   

   λe	= [h2 / 2meE]½
   	= [6.626 × 10-34 J-s)2 / (2)(9.11 × 10-11 kg)(6000 eV × 1.609 × 10-19 J/eV)]½
   	= 0.0158 nm

   

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10. Suppose that the velocity of an electron has been measured to within an uncertainty of ±1 cm/s. What minimum uncertainty in the position of the electron does this imply?

    The uncertainty principle is
    

    Δx Δp ³ h/4π .

    We are looking for Δx, so we first need to know Δp. Since p = mv, Δp = mΔv, hence

    Δx	³ h / 4pDp
    	³ h / 4πmΔv
    	³ (6.626 × 10-34 J-s) / (4)(9.11 × 10-31 kg)(1 × 10-2 m/s)
    	³ 6 × 10-3 m

    When you recall that atoms are a few nanometres wide, you realize that this is a huge volume of space for such a small particle.

    

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11. If the position of an automobile of mass 2 × 10³ kg is uncertain by ±10^-8 m, what is the corresponding uncertainty in its velocity?

    The uncertainty principle is
    

    Δx Δp ³ h/4π .

    We are looking for Δv, so we first need to know λp. Since p = mv, λp = mλv, hence

    Δv	³ h / 4πmΔx
    	³ (6.626 × 10-34 J-s) / (4)(2000 kg)(1 × 10-8 m)
    	³ 2 × 10-30 m/s

    Since Δx is about 10 atom widths is what we can actually measure the position of macroscopic objects using interferometry, we see that the uncertainty principle is unimportant for macroscopic objects.
    

    

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12. What are the orbital radii and electron velocities in the first three levels of the Bohr model of the hydrogen atom. (1/4pe₀ = 8.99 × 10⁹ N-m/C², e = 1.602 × 10^-19 C)
    

    In the Bohr model of the Hydrogen atom, the orbital radii are given by
    

    r_n = n²a₀ ,

    where a₀ = 0.0529 nm. The orbital velocities are given by
    

    v_n = h / 2πm_ea₀n .

    The results are

    n	rn (nm)	vn (m/s)
    1	0.0529	2.20 × 106
    2	0.2120	1.10 × 106
    3	0.4760	0.73 × 106

    

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13. Show for an ionized atom of charge Ze with only a single electron that the radius, speed, and energy of the electron in the n^th orbit are: r = n²a₀/Z, v = Ze²/(2nε₀h), E_n = (-13.6 eV)Z²/n². Note that the spectra for such an atom would have the same pattern as hydrogen but shifted in wavelength by Z².
    

    In this case the Coulomb force between the nucleus and the lone electron would be

    F_C = (1/4pe₀) (Ze)(e)/r² .

    The resulting electrostatic potential would be
    

    U_C = -òF_cdr = -(1/4pe₀) Ze²/r .

    Next we follow the steps in the derivation of the Bohr Model. Newton's Law for circular orbits, F_net = ma and a_centripetal = v²/r, yields
    

    (1/4pe0) Ze2/r2

    = mv2/r .

    (1)

    Quantizing the angular momentum of the orbit yields
    

    L

    = mevr = nh/2π ,

    (2)

    where n is an integer. We now have two equations in two unknowns, v and r. Solving first for v yields
    

    vn

    = Ze2/ (2nε0h) .

    (3)

    Solving for r, yields
    

    rn

    = nh / 2πmevn = (n2/Z) (hε0 / πmee2) = n2a0/Z .

    (4)

    Now the energy or an orbital is the sum of the kinetic and potential energies

    En	= K + U
    	= ½me (vn)2 - (1/4pe0) Ze2/rn
    	= -[mee4/ 8(εoh)2](Z/n2)
    	= -(13.6 eV)(Z/n2)

    

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14. When a block of metal serving as a target in an X-ray tube is bombarded with a beam of fast electrons, it emits X-rays of wavelength 0.167 nm. Assuming that these X-rays arise in transitions from the first excited state to the ground state, identify the metal.

    We use the results from questions 13, since the inner electrons are relatively unaffected by the outer electrons.

    The ground state is n = 1 and the first excited state is n = 2, the energy difference is

    ΔE	= Ef - Ei
    	= (-13.6 eV) Z2 [1/(n0)2 - 1/(n1)2]
    	= (-13.6 eV) Z2 [1/12 - 1/22]
    	= (-10.2 eV) Z2

    Now all this energy goes into creating a single photon. The relationship between ΔE and λ is
    

    ΔE = hc/λ ,

    where hc is 1240 nm-eV in convenient units.
    

    Combining our results yields an equation for Z,
    

    Z	= [hc/λ (10.2 eV)]½
    	= [(1240 nm-eV) / (0.167 nm)(10.2 ev) ]½
    	= [728]½
    	= 27

    The element with Z = 27 is cobalt.
    

    

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15. How far away can you see a 100-W light bulb at night? Assume that the light emitted has a wavelength of 550 nm and the threshold for your eye to detect light is 25 photons per second passing through your pupil. The diameter of a dark-adapted pupil is about 8 mm.
    
    

    The light bulb emits photons and energy equally in all directions. The energy of a single photon is
    

    E_photon = hc/λ = 1240 nm-eV / 550 nm = 2.545 eV = 3.614 × 10^-19 J .

    Now each second, the lightbulb emits an energy
    

    E_lightbulb = 100 W 1 s = 100 J .

    So the number of photons emitted by the lightbulb each second is
    

    N = E_lightbulb / E_photon = 100 J / 3.614 × 10^-19 J = 2.767 × 10²⁰ .

    Now the pupil intercepts only a portion of these photons proportional to the area or the pupil compared to the surface area of the sphere of radius R,
    

    N_eye = N A_eye/A_sphere = N ¼πd² / 4πR² = Nd²/16R² .

    Since we are told N_eye and d, the diameter of the pupil, we can find R
    

    R = ¼d[N/N_eye]^½ = ¼(8 mm) [2.767 × 10²⁰ / 25]^½ = 6.7 × 10⁶ m .

    This undoubtedly would require very good viewing conditions and no nearby light sources.
    

    

Questions? mike.coombes@kwantlen.ca