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Consider a Bohr model of doubly ionized lithium. (a) Write an expression similar to Equation 28.14 for the energy levels of the sole remaining electron. (b) Find the energy corresponding to $n=4 .$ (c) Find the energy corresponding to $n=2 .$ (d) Calculate the energy of the photon

emitted when the electron transits from the fourth energy level to the second energy level. Express the answer both in electron volts and in joules. (e) Find the frequency and wavelength of the emitted photon. (f) In what part of the spectrum is the emitted light?

a) $E_{n}=\frac{-122 \mathrm{eV}}{n^{2}}$

b) $-7.63 \mathrm{cV}$

c) $-30.5 \mathrm{cV}$

d) $3.67 \times 10^{-18} J$ or $ 22.95 \, \rm eV $

e) $\lambda = 54 \rm nm $, $ f = 5.52 \times 10^{15} \mathrm{Hz}$

f) ultraviolet

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In this exercise, we have the doubly ionized lithium Adam and in question A. We have to find the energy levels e n of this Adam. Remember that for ah, hydrogen like Adam E N The the energy of the ends energy level is given by minus disease square, divided by n square times 13.6 electoral votes in this case deliver him the doubly ionized. Leave him is a hydrogen like Adam because there's only one electrons orbiting the nucleus and Z, the atomic number of the lithium is equal to three eso. Yeah, for the leave him is his miner's nine, which is three square times, 13.6, divided by and square. So this is equal to minors. 122.4 electoral votes. If I could buy and square question be, we have to calculate explicitly the energy of the fourth energy level. So that's before, which is minus 122.4. If I didn't buy four square. So this is minus seven 0.65 electoral votes in question. See, we have to calculate the energy of the second energy level, so that's mine is 122.4 divided by two squared, which is for so this is minus 30.6 Acting votes in Question D We have to calculate the energy of affording that's emitted after the transition off the Adam from the fourth off the electorate in the Adam from the fourth enter your level to the 2nd 1 So the the elections of transitions from an Nichols for two and equals two and from conservation of energy, we have that the energy of the Fulton Gamma is equal to the initial energy. Mine is the final energy. So this is the four minus he to This is minus 7.65 plus 30.6 electoral votes, which is equal to 22.96 95 electoral votes. This is the answer in electoral votes and the answer in order to obtain the answer in Jews, we need to multiply this by 1.6 times 10 to the minus 19. You spare like to vote. So we get that the energy of a photon ISS 3.6 seven to times 10 to the miners 18th Use okay in question e, we have to find both the wavelength and the frequency. Ah, the fulton. So we know that the energy of a photon is H f. Or you can also be written as H C over London. So f is he divided by Age E is 22.95 electoral votes. H is 4.14 time since the minus 15 electoral votes. Second, so F is equal to 5.54 times 10 to them, 15 hurt The wavelength Lambda is, uh, h c divided by E h C stuff 100 and 40 election votes, then the meters and E is 22.95 electoral votes. So Lunda is equal to 54 millimeters. Finally, in the question f we have to to say in which part of the electromagnetic spectrum this photo is in, that is the the wavelength. London equals to 4 to 50 54 centimeters is in what part of the electromagnetic spectrum I noticed that the ultraviolet rays, the ultraviolet part of the spectrum, lies in a range between 10 centimeters until about 400 a meters. She air wavelength blunder lies within this range and we can say that the Fulton is in the ultraviolet part of this back

Universidade de Sao Paulo