CHAPTER 30 THE NATURE OF THE ATOM PROBLEMS ______________________________________________________________________________ 7. SSM WWW REASONING AND SOLUTION The radii for Bohr orbits are given by Equation 30.10: 2 n 11 rn = (5.29 ×10 m) Z 2+ are given by Equation 30.10 with The Bohr radii for a doubly ionized lithium atom Li Z 3; therefore, the radius of the n 5 Bohr orbit in a doubly ionized lithium atom Li 2+ is 2 5 10 = 4.41×10 m 3 ______________________________________________________________________________ r5 = (5.29 ×10 11 m) 23. SSM WWW REASONING The values that " can have depend on the value of n, and only the following integers are allowed: " 0, 1, 2, . . . (n – 1). The values that m " can have depend on the value of " , with only the following positive and negative integers being permitted: m" – ", . . . –2, –1, 0, +1, +2, . . .+" . SOLUTION Thus, when n 6, the possible values of " are 0, 1, 2, 3, 4, 5. Now when m" 2 , the possible values of " are 2, 3, 4, 5, . . . These two series of integers overlap for the integers 2, 3, 4, and 5. Therefore, the possible values for the orbital quantum number " that this electron could have are " = 2, 3, 4, 5 . 27. SSM WWW REASONING Let T denote the angle between the angular momentum L and its z-component Lz . We can see from the figure at the right that Lz L cos T . Using Equation 30.16 for Lz and Equation 30.15 for L, we have cos T Lz L m" h / 2S " " 1 ª¬ h / 2S º¼ +z Lz T L m" " " 1 The smallest value for T corresponds to the largest value of cos T. For a given value of " , the largest value for cos T corresponds to the largest value for m " . But the largest possible m " " . Therefore, we find that m " is value for Chapter 30 Problems cos T " " " 1 1071 " " 1 " . When the SOLUTION The smallest value for T corresponds to the largest value for electron is in the n 5 state, the largest allowed value of " is " 4 ; therefore, we see that " 4 T cos –1 4 / 5 or 26.6q "+1 4 1 ______________________________________________________________________________ cos T 41. SSM WWW REASONING The number of photons emitted by the laser will be equal to the total energy carried in the beam divided by the energy per photon. SOLUTION The total energy carried in the beam is, from the definition of power, E total Pt (1.5 W)(0.050 s) = 0.075 J The energy of a single photon is given by Equations 29.2 and 16.1 as E photon hf hc O (6.63 u 10 –34 J s)(3.00 u 10 8 m/s) 514 u 10 –9 m 3.87 u 10 –19 J –9 where we have used the fact that 514 nm = 514 u 10 m . Therefore, the number of photons emitted by the laser is E total 0.075 J 1.9 u 10 17 photons E photon 3.87 u 10 –19 J/photon ______________________________________________________________________________
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