1. Sodium atoms have a characteristic yellow color when excited in a flame. The color comes from the emission of light if 589.0 nm (this is the same experiment as the flame emission spectroscopy we did in lab 4 with potassium).
What is the frequency of this radiation? What is the energy of this radiation per photon? Per mole of photons?
ν = c/λ = 2.9979*108 m/s / 5.890*10-7 m = 5.090*1014 s-1 (s-1 is also called Hz – Hertz – and is the SI unit for frequency)
ΔE = hν = 6.626*1034 J*s * 5.090*1014 s-1 = 3.373*10-19 J – This is the energy per photon. To get ΔE per mole, multiply by Avogadro’s number to get 2.031*105 J/mol.
2. What is the wavelength (in nm) of an electron (mass = 9.11*10-31 kg) traveling at 5.31*106 m/s? What is the wavelength (in m) of a tennis ball (mass = 56 g) traveling at 80 km/hr? What does this tell you about quantum versus classical mechanics?
λ = h/m*v
For the electron: λ = 6.626*10-34 J*s /(9.11*1031 kg * 5.31*106 m/s) = 1.37*10-10 m = 0.137 nm
For the tennis ball: λ = 6.626*10-34 J*s /(0.56 kg * 22.22 m/s) = 5.324*10-35 m
For the electron: λ = 6.626*10-34 J*s /(9.11*1031 kg * 5.31*106 m/s) = 1.37*10-10 m = 0.137 nm
For the tennis ball: λ = 6.626*10-34 J*s /(0.56 kg * 22.22 m/s) = 5.324*10-35 m
Because the wavelength is so small for the tennis ball, it is not something that could be measured. This is why we use different equations for quantum mechanics and classical mechanics.
3. Calculate the change in energy corresponding to the relaxation of an electron from the n=4 state to the n=2 state in the hydrogen atom. What does the sign of this energy change tell you?
ΔE = Efinal – Einitial = E4 – E2 = -2.178*10-18 J [Z2/(42-22)] = -1.815*10-19 J (Z=1 here because we are dealing with the hydrogen atom). The sign tells you that energy is released in this process.
4. For the n=4 energy level, how many electrons can fit, including spin? What are the possible l values and to what orbitals do they correspond? What are the ml values for each l? What do the ml values tell you?
For n=4, there are s, p, d, and f orbitals which can fit 2, 6, 10, and 14 electrons, respectively. Therefore, 32 electrons can fit in n=4. The possible l values are 1, 2, 3, and 4, corresponding to the s, p, d, and f orbitals, respectively. The ml value for s is 0; for p, -1, 0, 1; for d, -2, -1, 0, 1, and 2; for f, -3, -2, -1, 0, 1, 2, 3. These values are related to the position of the orbital in space relative to the other orbitals. They are a way to label each orbital. Each orbital contains two spin-orbitals, which means that each orbital can contain an electron with spin +1/2 and spin -1/2.
5. What is the electron configuration of chromium? What is the configuration in shorthand notation? How many valence electrons are there? Load the orbital diagram with electrons.
Cr: 1s22s22p63s23p63d54s1 or in shorthand: [Ar] 3d54s1
There are 6 valence electrons.
When you draw the orbital diagram, be sure that all orbitals that are half full have electrons with the same spin. It is convention to draw the electrons in orbitals that are half full with spin up.
Note that the d orbital and the s orbital are both half full, rather than the d orbital having six electrons and the s orbital having none. Even though the 4s orbital is lower in energy than the 3d orbital, there is a stabilization energy associated with half-full orbitals.
When you draw the orbital diagram, be sure that all orbitals that are half full have electrons with the same spin. It is convention to draw the electrons in orbitals that are half full with spin up.
Note that the d orbital and the s orbital are both half full, rather than the d orbital having six electrons and the s orbital having none. Even though the 4s orbital is lower in energy than the 3d orbital, there is a stabilization energy associated with half-full orbitals.
6. According to periodic trends, order the following atoms in terms of atomic radius, ionization energy, and electron affinity.
a.) Li, B, N, F
a.) Li, B, N, F
Atomic radius: Li>B>N>F
Ionization energy: F>N>B>Li
Electron affinity: F>N>B>Li
b.) Fr, Sr, Ga, P
b.) Fr, Sr, Ga, P
Atomic radius: Fr>Sr>Ga>P
Ionization energy: P>Ga>Sr>Fr
Electron affinity: P>Ga>Sr>Fr
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