Use l 136 1240 n 1 n 1 2 2 2 1 find integers n 1 and

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Use Equs. 37-18 and 37-15; l 13.6 1240 = n 1 n 1 2 2 2 1 - Find integers n 1 and n 2 . Note n 1 must be 1 in this case. 0.9375 = n 1 n 1 2 2 2 1 - n 1 = 1, n 2 = 4. The transition is from n = 4 to n = 1. 68 · The wavelength of a spectral line of hydrogen is 1093.8 nm. Identify the transition that results in this line. Proceed as in Problem 67 Find n 1 and n 2 by trial and error 1/ n 1 2 – 1/ n 2 2 = 0.0833 n 1 = 3, n 2 = 6. The transition is from n = 6 to n = 3 69*· Spectral lines of the following wavelengths are emitted by singly ionized helium: 164 nm, 230.6 nm, and 541 nm. Identify the transitions that result in these spectral lines. 1. Express E n for He + ; note that Z = 2 2. Use Equs. 37-18 and 37-15 to write n 1 and n 2 in terms of ? 3. By trial and error identify the integers n 1 and n 2 for which ? = 164 nm, 230.6 nm, and 541 nm. E n = –(4 × 13.6/ n 2 ) eV = –54.4/ n 2 eV l 54.4 1240 = n 1 n 1 2 2 2 1 - For ? = 164 nm, n 2 = 3, n 1 = 2; for ? = 230.6 nm, n 2 = 9, n 1 = 3; for ? = 541 nm, n 2 = 7, n 1 = 4 70 ·· We are often interested in finding the quantity ke 2 / r in electron volts when r is given in nanometers. Show that ke 2 = 1.44 eV nm. From Equ. 37-16 we have ke 2 / a 0 = 27.2 eV, where a 0 = 0.0529 nm. So ke 2 = 1.44 eV . nm. 71 ·· The wavelengths of the photons emitted by potassium corresponding to transitions from the 4P 3/2 and 4P 1/2 states to the ground state are 766.41 nm and 769.90 nm. ( a ) Calculate the energies of these photons in electron volts. ( b ) The difference in the energies of these photons equals the difference in energy ? E between the 4P 3/2 and 4P 1/2 states in potassium. Calculate ? E . ( c ) Estimate the magnetic field that the 4p electron in potassium experiences. ( a ) hf = (1240/ ? ) eV; ? in nm hf = 1.6179 eV for ? =766.41 nm;
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