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4. (From Eisberg & Resnick, P 4-33, pg 122) Using Bohrâs model, calculate the energy required to remove the electron from singly ionized helium. [5] Solution: This is equivalent to finding the ground state energy of singly ionized helium, and taking the absolute value of that. We want to find the energy needed for a transition from the n = 1 to the n = â state, with Z = 2. En = â 1 4Ï0 2 me4 Z 2 2~2 1 n2 . 2 1 me4 Z 2 1 1 â 4Ï0 2~2 12 â2 2 1 me4 22 1 =â = â(4)(13.6 eV) 4Ï0 2~2 12 âE = E1 â Eâ = â = â(54.4 eV) = â(8.715 . . . Ã 10â20 J) It will take about 54.4 eV or 8.72 Ã 10â20 J of energy to remove an electron from singly ionized helium. 5. (From Eisberg & Resnick, P 4-35, pg 122) A 3.00 eV electron is captured by a bare nucleus of helium. If a 2400 AÌ photon is emitted, into what level was the electron captured? [5] Solution: Energy is conserved in this interaction. The initial energy is just the given kinetic energy of the electron, Ei = Ki = 3.00 eV, since the potential energy of atom is zero when the electron is far away from the nucleus. The final energy of the system is made up of the energy of the photon Ephoton = hc/Î» plus the energy of the electron in the n = 1 state in the helium atom (Z = 2). En = â Ephoton = 1 4Ï0 2 me4 Z 2 2~2 1 n2 = â(4)(13.6 eV) 1 n2 = â(54.4 eV) 1 n2 hc (4.1356668 . . . Ã 10â15 eV Â· s)(299792458 m/s) = = (5.166 . . . eV) Î» (2400 Ã 10â10 m) Ef = Ei En + Ephoton = Ki â(54.4 eV) En = Ki â Ephoton 1 = (3.00 eV) â (5.166 . . . eV) n2 (2.166 . . . ) 1 = = 0.039816 . . . 2 n (54.4) n2 = 25.11 . . . n=5 The electron was captured into the n = 5 level by the bare nucleus of helium.