Homework Solutions – Chapter 31 3. Picture the Problem: An alpha particle of charge 2e is brought from infinity to the surface of a copper nucleus. Strategy: Let the initial configuration correspond to the alpha particle at rest and infinitely far from the nucleus, and let the final configuration correspond to the alpha particle at rest at a distance of one nuclear radius. The work required to bring the alpha particle near the nucleus is the nonconservative work nc f iWEE=− (equation 8-9), but because the kinetic energy is zero in both configurations, nc.WU=Δ Use equation 20-8 to solve for the change in potential energy. Solution: 1. (a) Find the work required to bring an alpha particle from infinity to a distance r: ( )( )20nc229580ke ekq qkeWUrrr⎛⎞=Δ = − = =⎜⎟⎝⎠ 2. Insert the constants and set the distance equal to half the diameter of the nucleus: ( )( )( )2922 –19–151258 8.99 10 N m C 1.60 10 C5.6 pJ4.8 10 mW×⋅ ×==× Insight: The energy may seem small, but it is equivalent to 35 MeV, over 68 times the rest energy of an electron! 8. Picture the Problem: The image shows the energy levels for the three longest wavelength transitions in the Paschen series. Strategy: Use equation 31-2 to calculate the appropriate wavelengths. For the Paschen series set n’ = 3. The longest wavelengths correspond to the smallest values of n. Therefore, set n equal to 4, 5, and 6 to obtain the longest wavelengths. Solution: 1. Set n’ = 3 in eq. 31-2 and solve for the wavelength: 221113Rnλ=⎛⎞−⎜⎟⎝⎠ 2. Set n = 4: ( )4712211875 nm111.097 10 m34λ−==⎛⎞×−⎜⎟⎝⎠ 3. Set n = 5: ( )5712211282 nm111.097 10 m35λ−==⎛⎞×−⎜⎟⎝⎠ 4. Set n = 6: ( )6712211094 nm111.097 10 m36λ−==⎛⎞×−⎜⎟⎝⎠ Insight: All three of these wavelengths lie in the infrared portion of the electromagnetic spectrum. 11. Picture the Problem: Suppose the mass of the electron were magically doubled. Strategy: The ionization energy of hydrogen depends upon the energy of the n = 1 orbit. Equation 31-8 indicates that the magnitude of that energy is linearly proportional to the mass of the electron. This result follows from equation 31-5, where we see that the radius of an electron’s orbit is inversely proportional to the mass, and from equation 31-6, where we see that the speed of the electron is independent of its mass. Use these relationships to answer the question. Solution: 1. (a) Referring to equation 31-8, we can see that the total (negative) energy of the electron would double with its mass, as would the energy required to ionize the atom. We conclude that if the mass of the electron were magically doubled, the ionization energy of hydrogen would increase. 2. (b) The best explanation is I. The ionization energy would increase because the increased mass would mean the electron would orbit closer to the nucleus and would require more energy to move to infinity. Statements II and IIIare each false. Insight: Statement III is false because the force holding the electron in orbit is the electrical force, not the gravitational force, so the mass does not cancel out of the equations.12. Picture the Problem: Three atoms or ions have a single electron orbiting the nucleus: (A) neutral hydrogen in the state n = 2; (B) singly ionized helium in the state n = 1; and (C) doubly ionized lithium in the state n = 3. Strategy: The Bohr model can be used to describe atoms or ions that have a single electron. Use the expression 22224nhrnmk Zeπ⎛⎞=⎜⎟⎝⎠ (equation 31-7) to determine the ranking of the Bohr radii. Solution: Noting that the radius of Bohr orbits is proportional to 2,nZwe find that for atom (A), n = 2 and Z = 1, so 2221 4.nZ== For atom (B) 2212 12,nZ== and for atom (C) 2233 3.nZ== We conclude that B < C < A. Insight: For each energy state n the orbit radii get progressively smaller as Z increases because the electron is more tightly bound to the nucleus as the amount of positive charge in the nucleus increases. 14. Picture the Problem: An electron in the n = 1 Bohr orbit has the kinetic energy K1. Strategy: Use the expressions in the derivation leading to equation 31-8 to note that 2.2keZEKr=− =− Use this relationship and the fact that 21nEn∝ to determine the relationship between K2 and K1. Solution: We can see from the derivation leading up to equation 31–8 that the kinetic energy of a Bohr orbit is equal to minus one times the total energy of the orbit; that is, .KE=−In addition, recall that E is proportional to 21.n Therefore, the kinetic energy of an electron in the n = 2 Bohr orbit is 1214.KK= Insight: We can also find the answer by means of a ratio: ( )( )222222 12122111213.6 eV11 1 24 413.6 eVnKE nKKKEnn−⎛⎞== == =⇒ =⎜⎟−⎝⎠ 18. Picture the Problem: The energy of the electron is determined by its Bohr orbit. To switch between orbits, a photon with energy equal to the difference in orbital energies must be absorbed by the electron. Strategy: Set the energy of the absorbed photon equal to the change in electron energies between the n = 2 and n = 6 orbits. The orbit energies are given by equation 31-9. Solution: Use equation 31-9 to find :EΔ ( )52221113.6 eV 3.02 eV62EE E⎛⎞Δ= − =− − =⎜⎟⎝⎠ Insight: This energy corresponds to a photon of wavelength 410 nm, in the violet portion of the visible spectrum. 22. Picture the Problem: When energy is added to an electron in the n = 3 orbit, it will jump to a higher orbit. We want to calculate the final energy level attained when 1.23 eV is added to such an electron. Strategy: Calculate the energy in the n = 3 orbit using equation 31-9 (with Z = 1). Add the 1.23 eV to the result to calculate the energy of the final orbit. Then use equation 31-9 to find the value of n in the final state of the electron. Solution: 1. Calculate the energy of the n = 3 orbit: 3213.6 eV1.51 eV3E =− =− 2. Calculate the final energy level: 31.51 eV 1.23 eV = 0.281 eVnEE E=+Δ=− + − 3. Solve for the quantum number n: 213.6 eV 13.6 eV 13.6 eV 70.281 eVnnEnEn−−=− ⇒ = = =− Insight: If the electron were to now spontaneously drop back down to its ground state, it would give off an ultraviolet photon that has energy ( )0.281 e V 13.6 eV 13.3 eVEΔ=− −− =
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