HW04-key - Chemistry 111 Homework#4 KEY Due(Wednesday 1...

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Chemistry 111 Homework #4 – KEY Due: September 28, 2004 (Wednesday) 1. What is the radius of the Be 3+ ion in its n=6 level in the Bohr model? (a) 4.76 Å (b) 4.76 nm (c) 0.794 Å (d) 5.88 pm (e) 5.88 Å (f) not in the list In the Bohr model … 22 2 0 0 2 e nh ε n r= = a Ze m π Z Å recall, a 0 = Bohr radius = 0.529 Å = (6 2 / 4) (0.529 Å) = 4.76 Å The next 3 questions are True – False, and relate to problem 1 above. 2. If the Be 3+ ion above absorbs a photon, its radius will decrease. The only way the Be 3+ ion could absorb a photon is for its electron to use the energy to jump from the n=6 level to a higher level. The radius increases with n 2 , so the answer is False. 3. The radius of the Li 2+ ion in its n=6 level is 1.333 times larger than the radius of the Be 3+ ion above. The radius is inversely proportional to the atomic number Z. Since Z = 4 for Be 3+ , and Z = 3 for Li 2+ , the radius of the Li 2+ ion is 4/3 times the radius of the Be 3+ ion in any level. True. 4. The radius of the Be 3+ ion above is equal to the radius of the C 5+ ion in its n=9 state. Now, both Z and n are changing, and in fact n/Z = 1.5 for both ions. However, we need to examine n 2 /Z for each of these atoms. For Be 3+ , n 2 /Z = (36/4) = 9. For C 5+ , n 2 /Z = (81/6) = 15.2. The C 5+ ion in n=9 is larger than the Be 3+ in n=6 The answer is thus False.
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Chemistry 111 2 Homework #4 / Fall 2005 Answer Key 5. What is the energy of the Be 3+ ion in its n=6 state? (a) -1.92 x 10 -17 J (b) 9.69 x 10 -19 J (c) -2.18 x 10 -18 J (d) -9.69 x 10 -19 J (e) -4.91 x 10 -18 J (f) not in the list Hmmm … now we need to calculate the energy of this ion. 24 2 e 0 222 2 0 Zem Z E= = E 8n h ε n −− Å E 0 = 2.180 x 10 -18 J E = -2.180 x 10 -18 J (4 2 /6 2 ) = - 9.689 x 10 -19 J 6. The Be 3+ ion above drops down to the n=2 level. How many of the following statements are TRUE ? 3 The transition is in the Be 3+ equivalent of the Balmer series of the hydrogen spectrum. True. The Balmer series of the hydrogen atom is composed of transitions from n > 2 levels to the n=2 level. A photon of 25.6 nm is emitted. True. Well, a photon is certainly emitted during this process. Now we need to calculate its wavelength. We can use the following equation: Δ E = -2.180 x 10 -18 J (Z 2 ) (1/2 2 – 1/6 2 ) = 7.751 x 10 -18 J Δ E = h c / λ Æ λ = h c / Δ E = (6.626 x 10 -34 J s) (2.998 x 10 8 m s -1 ) / 7.751 x 10 -18 J = 2.56 x 10 -8 m = 25.6 nm
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Chemistry 111 3 Homework #4 / Fall 2005 Answer Key A photon of 25.6 nm is absorbed. False. Clearly this is false – moving to a smaller value of n (n=6 Æ n=2) results in the emission of a photon, not absorption. The radius of this ion decreases. True. Moving to a smaller value of n (n = 6 Æ n = 2) moves the electron closer to the nucleus. The radius therefore decreases. The photon emitted or absorbed has a longer wavelength than the photon emitted or absorbed when C 5+ undergoes the same n=6 Æ n=2 transition. False. Hmmm … once again, we need to look at the relevant equation to make a prediction. The relevant equation is Δ E = -2.180 x 10 -18 J (Z 2 ) (1/2 2 – 1/6 2 ) Z C > Z Be Æ the change in energy in going from n=6 Æ n = 2 is greater in C 5+ than in Be 4+ . Wavelength is inversely proportional to energy, so the wavelength should decrease.
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HW04-key - Chemistry 111 Homework#4 KEY Due(Wednesday 1...

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