Your Turn! WCBS broadcasts at a frequency of 880 kHz. What is the wavelength of their signal? 341 m 293 m 293 mm 341 km = 341 m Converting from frequency.

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Presentation transcript:

Your Turn! WCBS broadcasts at a frequency of 880 kHz. What is the wavelength of their signal? 341 m 293 m 293 mm 341 km = 341 m Converting from frequency to wavelength.

Calculate the wavelength of red light with a frequency of 4 Calculate the wavelength of red light with a frequency of 4.62 x 1014 s−1 Given: Find: n = 4.62 x 1014 s−1 l, (nm) Conceptual Plan: Relationships: l∙n = c, 1 nm = 10−9 m n (s−1) l (m) l (nm) Solve: Converting from frequency to wavelength. Check: the unit is correct, the wavelength is appropriate for red light

Determine the wavelength (in nm) of an X-ray with a frequency of 4 Determine the wavelength (in nm) of an X-ray with a frequency of 4.2 × 1018 Hz. 7.1 × 10–11 7.1 × 10–2 1.3 × 1027 1.4 × 1010 7.1 × 10–18 Converting from frequency to wavelength. Wavelength = c/frequency

Determine the wavelength (in nm) of an X-ray with a frequency of 4 Determine the wavelength (in nm) of an X-ray with a frequency of 4.2 × 1018 Hz. 7.1 × 10–11 7.1 × 10–2 1.3 × 1027 1.4 × 1010 7.1 × 10–18 Answer: b

Learning Check What is the frequency, in sec–1, of radiation which has an energy of 3.371 × 10–19 joules per photon? Energy converted to frequency E=hv  = 5.087 × 1014 s–1

Your Turn! A microwave oven uses radiation with a frequency of 2450 MHz (megahertz, 106 s–1) to warm up food. What is the energy of such photons in joules? 1.62 × 10–30 J 3.70 × 1042 J 3.70 × 1036 J 1.62 × 1044 J 1.62 × 10–24 J Frequency converted to energy.

What is the wavelength of a particle that has an energy of 4 What is the wavelength of a particle that has an energy of 4.41 × 10–19 J? 441 nm 450 nm 227 nm 222 nm 199 nm Energy converted to wavelength E = hc/wavelength

What is the wavelength of a particle that has an energy of 4 What is the wavelength of a particle that has an energy of 4.41 × 10–19 J? 441 nm 450 nm 227 nm 222 nm 199 nm Answer: b wavelength = hc/E

Ultraviolet light emits a total of 2 Ultraviolet light emits a total of 2.5 × 10–17 J of light at a wavelength of 9.8 × 10–7 m. How many photons does this correspond to? 1 10 25 100 125 A little more complicated problem. Remember total energy corresponds to all of the photons. But each photon has a characteristic wavelength or frequency. Focus on the wavelength in this problem. What is the energy of a photon with that wavelength? Once you know that you can calculate the number of photons needed to get the total energy. Energy per photon = hc/wavelength # photons = energy/ energy per photon

Ultraviolet light emits a total of 2 Ultraviolet light emits a total of 2.5 × 10–17 J of light at a wavelength of 9.8 × 10–7 m. How many photons does this correspond to? 1 10 25 100 125 Answer: e Energy per photon = hc/wavelength # photons = energy/ energy per photon

Calculate the number of photons in a laser pulse with wavelength 337 nm and total energy 3.83 mJ. Given: Find: l = 337 nm, Epulse = 3.83 mJ number of photons Conceptual Plan: Relationships: E=hc/l, 1 nm = 10−9 m, 1 mJ = 10−3 J, Etotal=Ephoton  # photons l(nm) l (m) Ephoton number photons Solve: This problem is the same as the last one. Calculate the energy of a single photon at this wavelength. Then realize the total energy is contributed by many photons. In these problems, you must be careful of your units – they often require converting meters to nanometers. Here you have to worry about millijoules. How many joules is that?

Learning Check: Using Rydberg Equation Consider the Balmer series where n1 = 2 Calculate  (in nm) for the transition from n2 = 6 down to n1 = 2. = 24,373.9 cm–1 Note – The Balmer series is the four lines of hydrogen in the visible spectrum. They correspond to n1 being equal to 2. There are other series of lines but they are not in the visible spectrum. Here we are simply plugging in values into the Rydberg equation to calculate a wavelength. Make sure of your units! Here we are going from cm to nm.  = 410.292 nm Violet line in spectrum

Your Turn! What is the wavelength of light (in nm) that is emitted when an excited electron in the hydrogen atom falls from n = 5 to n = 3? 1.28216 × 103 nm 1.46232 × 104 nm 7.80002 × 102 nm 7.80002 × 10–4 nm 3.65341 × 10–7 nm Another simple exercise in plugging in values to calculate wavelength. Here n1 is equal to 3. This series is in the infrared region of the spectrum and not visible to the naked eye.

Learning Check A photon undergoes a transition from nhigher down to n = 2 and the emitted light has a wavelength of 650.5 nm. What is the value for nhigher? Just a slightly more complicated problem. Here they ask you to calculate n2 given the wavelength and the fact that n1 is equal to 2. Take care here. Nanometers and centimeters in the same equation! The units must be the same. You can use either nanometers or centimeters but you can’t mix both! n2 = 3

Your Turn! In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus, energy is emitted energy is absorbed no change in energy occurs light is emitted none of these

Your Turn! While the Bohr modeled failed to predict the line spectra of other atoms, it is historically important because it was Niels Bohr’s only major contribution to science it introduced the existence of neutrons it provided a way to measure the mass of electrons it introduced the idea of quantized energy for electrons in orbitals none of these

Your Turn! Calculate the de Broglie wavelength of a baseball with a mass of 0.10 kg and traveling at a velocity of 35 m/s. 1.9 × 10–35 m 6.6 × 10–33 m 1.9 × 10–34 m 2.3 × 10–33 m 2.3 × 10–31 m l = 1.9  10-34 m

How many orbitals have n = 3 and ml = –1? 2 3 4 5 For n = 3, l can be 2, 1, or 0 But ml cannot = -1 for l = 0. So the correct answer is 2.

How many orbitals have n = 3 and ml = –1? 2 3 4 5 Answer: b For n = 3, l can be 2, 1, or 0 But ml cannot = -1 for l = 0. It can only = -1 for the case where l =1 or l=2. The correct answer is 2.

What are the quantum numbers and names (for example, 2s, 2p) of the orbitals in the n = 4 principal level? How many orbitals exist? Given: Find: n = 4 orbital designations, number of orbitals Conceptual Plan: Relationships: l: 0 → (n − 1); ml: −l → +l n l ml 0 → (n − 1) −l → +l Solve: n = 4  l : 0, 1, 2, 3 n = 4, l = 0 (s) ml : 0 1 orbital 4s n = 4, l = 1 (p) ml : −1,0,+1 3 orbitals 4p n = 4, l = 2 (d) ml : −2,−1,0,+1,+2 5 orbitals 4d n = 4, l = 3 (f) ml : −3,−2,−1,0,+1,+2,+3 7 orbitals 4f total of 16 orbitals: 1 + 3 + 5 + 7 = 42 : n2

Which of the following transitions for an electron in a hydrogen atom would release the largest quantum of energy? n = 3 → n = 1 n = 4 → n = 3 n = 1 → n = 4 n = 2 → n = 1

Which of the following transitions for an electron in a hydrogen atom would release the largest quantum of energy? n = 3 → n = 1 n = 4 → n = 3 n = 1 → n = 4 n = 2 → n = 1 Answer: a

Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 6 to n = 5 Given: Find: ni = 6, nf = 5 l, m Conceptual Plan: Relationships: E = hc/l, En = −2.18 x 10−18 J (1/n2) ni, nf DEatom Ephoton l DEatom = −Ephoton Solve: Ephoton = −(−2.6644 x 10−20 J) = 2.6644 x 10−20 J Note – you could solve this problem by calculating the energy levels for n = 5 and n = 6 separately and then subtracting the initial level from the final level to get the energy change. At that point E = hc/wavelength would solve the problem. One could also use the Rydberg equation directly to solve the problem. Check: the unit is correct, the wavelength is in the infrared, which is appropriate because it is less energy than 4→2 (in the visible)

Example: Quantum Numbers How many subshells are in the fourth shell (n=4)? How many orbitals are in each of these subshells?

Which of the following is NOT an allowed set of quantum numbers? n = 4 l = 3 ml = 3 n = 1 l = 0 ml = 0 n = 5 l = 4 ml = –2 n = 2 l = 1 ml = 0 n = 3 l = 3 ml = –2

Which of the following is NOT an allowed set of quantum numbers? n = 4 l = 3 ml = 3 n = 1 l = 0 ml = 0 n = 5 l = 4 ml = –2 n = 2 l = 1 ml = 0 n = 3 l = 3 ml = –2 Answer: e

What is wrong with the following set of quantum numbers What is wrong with the following set of quantum numbers? n = 2, l = 2, ml = 0, ms = +1/2 ml must be greater than or equal to l. l must be greater than or equal to n. l has a maximum value of n–1. ms must be 0 or 1. This is an allowed set of quantum numbers.

Which orbital is described by the following set of quantum numbers Which orbital is described by the following set of quantum numbers? n = 3 l = 1 ml = –1 1s 2s 3s 3p 3d

Which orbital is described by the following set of quantum numbers Which orbital is described by the following set of quantum numbers? n = 3 l = 1 ml = –1 1s 2s 3s 3p 3d Answer: d

ml must be greater than or equal to l. What is wrong with the following set of quantum numbers? n = 2, l = 2, ml = 0, ms = +1/2 ml must be greater than or equal to l. l must be greater than or equal to n. l has a maximum value of n–1. ms must be 0 or 1. This is an allowed set of quantum numbers. Answer: c

Your Turn! Which of the following is a valid set of four quantum numbers (n, ℓ, mℓ , ms)?  3, 2, 3, +½  3, 2, 1, 0  3, 0, 0, –½  3, 3, 0, +½  0, –1, 0, –½

Your Turn! What is the maximum number of electrons allowed in a set of 4p orbitals? 14 6 2 10

Your Turn! Determine the number of electrons in an atom that can have a principal quantum number of 3, and list the possible subshells they occupy. 18 (3s, 3p, 3d) 8 (3s, 3p) 2 (3s) 32 (3s, 3p, 3d, 3f)

Which of the following correctly illustrates the valence electron configuration of sulfur? b) Remember Hund’s Rule! c) d)

Which of the following correctly illustrates the valence electron configuration of sulfur? b) Answer: a c) d)

Which of the following is the correct electron configuration for bromine? 1s22s22p63s23p64s24d104p5 1s22s22p63s23p64s23d104p5 1s22s22p63s23p64s23d104p6 1s22s22p63s23p64s24p6 1s22s22p63s23p54s24p5

Which of the following is the correct electron configuration for bromine? 1s22s22p63s23p64s24d104p5 1s22s22p63s23p64s23d104p5 1s22s22p63s23p64s23d104p6 1s22s22p63s23p64s24p6 1s22s22p63s23p54s24p5 Answer: b

What is the ground state electron configuration for V? V: [Kr]4s33d2 V: [Kr]4s24d3 V: [Ar]4s23d3 V: [Ar]3d3

What is the ground state electron configuration for V? V: [Kr]4s33d2 V: [Kr]4s24d3 V: [Ar]4s23d3 V: [Ar]3d3 Answer: c

How many valence electrons does an atom of vanadium possess? 2 3 4 5

How many valence electrons does an atom of vanadium possess? 2 3 4 5 Answer: e For transition metals we count those electrons in the s orbital and also those in the d orbitals even though they do not have the same principle quantum number.

Which element’s 4+ ion would have the following electron configuration? Te Sn Se Sb [Kr]4d10

Which element’s 4+ ion would have the following electron configuration? Te Sn Se Sb [Kr]4d10 Answer: c

Which of the following is the correct electron configuration for the anion typically formed from oxygen? 1s22s22p4 1s22s22p2 1s22s22p6 1s22s22p3 1s22s22p5

Which of the following is the correct electron configuration for the anion typically formed from oxygen? 1s22s22p4 1s22s22p2 1s22s22p6 1s22s22p3 1s22s22p5 Answer: c