FINAL - Chapter 1 Atomic Structure

Question 1

Rubidium has a work function (φ) of 3.35 ×
10-19 J. What is the longest wavelength of light
which will cause electrons to be emitted?

Answer 1

The kinetic energy is 0 because longest wavelength gives just enough energy to overcome the work function, but no extra energy to provide the electron with kinetic energy

𝐾𝐸 = ℎ𝜈 − φ
ℎ𝜈 − φ = 0
𝜈 = φ/ℎ

solve for velocity
now solve for wavelength using l = c/v

Question 2

Interactions between particles and waves?

Answer 2

Particles through collisions and waves through interference

Question 3

Explain the photoelectric effect

Answer 3

If photon energy is greater than the work function, then the electron is ejected.

If it is equal, electron is ejected but with no kinetic energy

KE = hv - ϕ
where ϕ is the binding energy

Electrons are ejected from polished metal surfaces by shining light on surface with a frequency above threshold. Higher frequency means higher KE, and the number of electrons ejected depends on brightness of light

Question 4

What atoms does Rydberg equation work for

Answer 4

Hydrogen or hydrogen like ions

Question 5

What is a node? What is an atomic orbital

Answer 5

It is defined where 𝛹 = 0 AND |ψ|² = 0 due to radial or angular term.
wave function

Points on standing wave where there is zero probability of finding an electron

Atomic orbital is region of space where there is a high probability of finding an e

Question 6

Define ionization energy

Answer 6

Energy to move electron from lowest filled energy level to infinity

Question 7

For this equation:
∆𝐸 = −2.178 × 10−18𝑍^2 (1/n^2f - 1/n^2i) what do positive and negative ∆𝐸 values mean?

Answer 7

Positive means photon is absorbed, and negative means that photon is emitted

Question 8

What does the IE equaton work for?

Answer 8

Only hydrogen and hydrogen like atoms

Question 9

Ionization energy of Ar is 1527 kj/mol. Lowest possible frequency of light to ionize the atom?

WATCH UR UNITS POR FAVOR

Answer 9

You have to convert IE to simply J/mol by dividing by avogadro’s number.

Make sure you’re in correct units!!!!

Whenever using IE equation, should always be in J/atom

Question 10

Photons with a minimum energy of
𝟒𝟗𝟔 𝒌𝑱/𝒎𝒐𝒍 are required to ionize Na atoms.
If light of 𝟔𝟎𝟎 𝒌𝑱/𝒎𝒐𝒍 was used, what would be the velocity of
the emitted electrons?

Answer 10

You already used the minimum energy of 496 as the work function, so the leftover kinetic energy is 600-496 and you convert this to J/atom and use

KE = 1/2mv^2
you should get 6.17 x 10^5

Question 11

Outline the four quantum numbers

Answer 11

Principle Quantum Number (n)
Size of an atomic orbital
(Positive integers)

Angular Momentum Quantum Number (l)
Orbital shape (s, p, d, f)
Zero to n-1

Magnetic Moment (mL) Quantum Number (ms)
Orientation of orbital
-l, 0, +l

Spin Quantum Number (ms)
Spin of electron
1/2 (pos spin) or -1/2 (neg spin)

Question 12

Outline Pauli’s exclusion principle

Answer 12

No two particles in same quantum system can simultaneously have identical values for all 4 quantum numbers

That is why given orbital can only have 2e’s, each with different spins

Question 13

Aufbau’s Principle

Answer 13

electrons fill in order of lowest n+l sum or if the same sum, in order of lowerst n

Question 14

What does a higher orbital energy mean in terms of IE?

Answer 14

Smaller IE, closer to 0

Question 15

Explain why each e- removal increases Zeff

Answer 15

Each e- removal increases Zeff and IE becuase number of shielding electrons decreases, so remaining electrons feel stronger pull by nucleus

Question 16

Provide the four explanations for trends in ionization energy

Answer 16

1) Effective nuclear charge (only works for atoms in the same row)
- Greater Zeff means greater IE becuase nucleus pulls stronger on e- so more energy required to remove an e-
- Greater Zeff means less shielding

2) Energy levels (shells)
- Electrons in higher n experience greater shielding and are less tightly bound to the nucleus so they have less IE

Both 1/2 have to do with the relationship IE is proportional to Zeff^2/n^2

3) Orbitals (also only works for atoms in the same level)
- S orbital has less energy and closer to nucleus compared to p orbital
- Higher orbital energy means smaller IE
S has higher IE than P

4) Spin pairing energy (emphasize the word energy and not repulsion)
Due to electron configuration, electron repulsion. Those with electron repulsion have lower IE

Question 17

What assumptions are made when using the Bohr energy equation

En = -2.178x10^-18 (Zeff^2/n^2)
not bound energy eq

Answer 17

Bohr energy equation only works for hydrogen or hydrogen like atoms (nucleus and 1e)

All atoms can be approximated as hydrogen like if all electrons except the one being ionized are assumed to ONLy shield the ionized electron from nuclear charge (rather than shielding other electrons too)

Question 18

Outline what shielding constant is

Answer 18

How much the effect of nuclear charge is removed by other electrons in the atom

Question 19

How does nuclear charge and shielding constant relate to first IE?

Answer 19

Nuclear charge increases by 1 every time atomic number increases while shielding constant increases by less than 1 so increases effective nuclear charge which increases IE across row

Question 20

What is ground state?

Answer 20

Lowest possible energy configuration

Obeys Aufbau’s (filling), most stable

Question 21

How to find number of unpaired electrons?

Answer 21

write out electron configuration