Lesson 2: Atom Structure

Question 1

What is the purpose of a mass spectrometer?

Answer 1

To determine the mass of a given atom.

Question 2

What is the equation for magnetic force in terms of magnetic field strength and velocity of an ion?

Answer 2

F = qvB F = Magnetic Force q = Charge of ion V = Velocity of ion B = Magnetic Field Strength

Question 3

In a mass spectrometer, the ion moves in a circle. Why?

Answer 3

The ion moves in a circle because the magnetic force is perpendicular to the velocity of the ion, pointing toward the center of the circle. The magnetic force essentially acts as a centripetal force.

Question 4

Which equation can be used in tandem with F = qvB in order to calculate the mass of an ion in a mass spectrometer based on the radius of the circle?

Answer 4

Fc = m(v^2/r)

Question 5

What is the resulting equation when everything is set equal to r?

Answer 5

r = mv/qB

Question 6

Of the following ions, which will have the smallest radius in a mass spectrometer? (A) Carbon-12 (B) Nitrogen-14 (C) Oxygen-16 (D) Potassium-19

Answer 6

(A) Carbon-12 The smaller the mass, the smaller the radius of the circle.

Question 7

The photoelectric effect describes the phenomenon in which a(n) ________ strikes a metal, resulting in the emission of a(n) ________ (A) Electron, electron (B) Electron, photon (C) Photon, photon (D) Photon, electron

Answer 7

(D) Photon, electron The photoelectric effect describes the phenomenon in which a photon strikes a metal, resulting the emission of an electron.

Question 8

The energy of the photon is greater than, equal to, or less than the energy of the emitted electron? Why?

Answer 8

The energy of the photon is greater than the energy of the emitted electron. This is because some of the energy of the photon was required (and used up) in order to free the electron from the metallic surface.

Question 9

What is the work function?

Answer 9

The work function is the energy of the photon that was required (and used up) in order to free the electron from the metallic surface.

Question 10

What is the equation for the energy of a photon in terms of the energy of the emitted electron?

Answer 10

KEp = KEe + WF KEp = Kinetic energy of the photon KEe = Kinetic energy of the electron WF = Work function

Question 11

A metal has a work function equal to 3.42 ⋅ 10^-19 J and the energy of the electron is equal to 3.47 ⋅ 10^-19 J. What is the frequency of the photon? (remember Plank’s constant = 6.626 ⋅ 10^-34). (A) 3.04 ⋅ 10 ^ -19 Hz (B) 6.02 ⋅ 10 ^ -13 Hz (C) 8.06 ⋅ 10 ^ 12 Hz (D) 1.15 ⋅ 10 ^ 15 Hz

Answer 11

(D) 1.15 ⋅ 10 ^ 15 Hz KEp = KEe + WF KEp = 3.47 ⋅ 10^-19 J + 3.42 ⋅ 10^-19 J KEp = 6.89 ⋅ 10^-19 J hv = KEp (6.626 ⋅ 10^-34)v = 6.89 ⋅ 10^-19 J v = (7 ⋅ 10^-19)/(6 ⋅ 10^-34) v = about 1 ⋅ 10^15 hz (Actual 1.15 ⋅ 10^15 hz)

Question 12

Which physics equation relates the energy of a photon to its frequency?

Answer 12

E = hf E = Energy of a photon or quantum h = Planck’s Constant (6.626 ⋅ 10^-34 J⋅s) f = Frequency of wave/radiation

Question 13

What two equations can be used in concert with Fc = m(v^2/r) to determine the radius of a Bohr model electron?

Answer 13

Fe = k((q1q2)/r^2) and L = rp

Question 14

Based on these equations, what equation can be used to determine the radius of any electron in terms of its principal quantum number?

Answer 14

rn = n^2 r1 rn = Radius of electron at n n = Principle quantum number r1 = Radius of electron at n of 1 (5.3 ⋅ 10^-11)

Question 15

An electron’s total energy is composed of what two types of energy? (A) Thermal and kinetic energy (B) Kinetic and electric energy (C) Electric and potential energy (D) Thermal and electric energy

Answer 15

(B) Kinetic and electric energy An electron’s total energy is composed of kinetic and electric energy.

Question 16

What equation can be used to determine the energy of any electron in terms of its principal quantum number?

Answer 16

En = E1/(n^2) En = Energy of electron at n n = Principle quantum number E1 = Energy of electron at n=1 (-2.17 ⋅ 10^-18 J)

Question 17

Which of the following is the conversion factor from Joules to electron-Volts (eV)? (A) 1 eV = 6.626 ⋅ 10^-34 J (B) 1 eV = 1.6 ⋅ 10^-19 J (C) 1 eV = 3 ⋅ 10^8 J (D) 1 eV = 9.11 ⋅ 10^-31 J

Answer 17

(B) 1 eV = 1.6 ⋅ 10^-19 J (A) [6.626 ⋅ 10^-34]is equal to Planck’s constant (C) [3 ⋅ 10^8] is equal to the speed of light (D) [9.11 ⋅ 10^-31] is equal to the mass of an electron

Question 18

The energy at the first orbital is equal to -2.17 ⋅ 10^-18 J. What is the energy in terms of electron-volts (eV)? (A) 19.2 eV (B) 3.7 eV (C) -1.4 eV (D) -13.6 eV

Answer 18

(D) -13.6 eV (-2.17 ⋅ 10^-18 J) / (1.6 ⋅ 10^-19) = approximately -10 eV (actual: -13.6 eV)

Question 19

The energy at the first orbital is equal to -13.6 eV. What is the energy in terms of electron-volts (eV) of an electron in the third orbital? (A) -9.6 eV (B) -4.8 eV (C) -3.4 eV (D) -1.5 eV

Answer 19

(D) -1.5 eV En = E1/(n^2) E3 = -13.6 eV / 3^2 E3 = approx. -1.5 eV

Question 20

An electron drops from the second orbital to the first orbital. A photon with how much energy is given off? (A) -9.6 eV (B) 6.3 eV (C) 10.2 eV (D) 13.6 eV

Answer 20

(C) 10.2 eV En = E1/(n^2) E2 = -13.6 eV / 2^2 E2 = approx. -3 eV E1 - E2 = -13.6 eV - (-3eV) = -10.6 eV from the electron, so the photon has an energy of approx. 10.6 eV (actual: 10.2 eV)

Question 21

What does it mean in terms of energy levels of electrons to say that the ionization energy of an atom is -13.6 eV?

Answer 21

It would require 13.6 eV to move an electron from n=1 to n=infinity, making the Hydrogen atom an ion.

Question 22

Ernest Rutherford conducted an experiment in which he fired alpha particles at gold foil. He noticed that the majority of alpha particles were not deflected by the gold foil, and all alpha particles would exit the foil. What did these results reveal about an atom’s structure?

Answer 22

The Rutherford experiments concluded that an atom has a dense, positively charged nucleus taking up a small fraction of an atom’s volume.

Question 23

Which 20th century scientist’s theory built upon Rutherford’s experiments to develop a Hydrogen atom model, relying on a single electron revolving around a nucleus and only allowing for discrete energy changes between electrons? (A) Niels Bohr (B) Richard Feynman (C) Albert Einstein (D) John Dalton

Answer 23

(A) Niels Bohr The Bohr model of an atom relies on an electron revolving around a nucleus and only allowing for discrete energy changes between electrons.

Question 24

Based on the previous description of the Bohr model, which of the following atoms does not represent the Bohr model? (A) H (B) He+ (C) Li+ (D) Be 3+

Answer 24

(C) Li+ The Bohr model relies upon there being only a single electron. Li+ has two electrons, whereas all other options have only one electron.

Question 25

True or false? Based on the Bohr model, the energy of an orbiting electron is directly proportional to the principal quantum number, with larger principal quantum numbers having less energy.

Answer 25

False. Based on the Bohr model, the energy of an orbiting electron is directly proportional to the principal quantum number, with larger principal quantum numbers having MORE energy.

Question 26

True or false? The Rydberg unit of energy is equal to 2.18 ⋅ 10 ^-18 J/electron, and is the experimentally determined energy of an electron at the smallest possible orbital.

Answer 26

True. The Rydberg unit of energy is equal to 2.18 ⋅ 10 ^-18 J/electron, and is the experimentally determined energy of an electron at the smallest possible orbital.

Question 27

Which of the following is not a characteristic of an electron moving from a lower energy orbit to a higher energy orbit? (A) Emitting light (B) Increased distance from the nucleus (C) Excitation of the electron (D) Increasing potential

Answer 27

(A) Emitting light An electron moving from a lower energy orbit to a higher energy orbit will ABSORB light, have higher potential, be excited, and increase its distance from the nucleus. Think of the mnemonic “AHED” [Absorb, High potential, Excited, Distant from nucleus] for this!

Question 28

An atom in the _________ state will emit a photon to release energy and return to the ________ state. (A) Ground, excited (B) Excited, ground (C) Elevated, resting (D) Resting, elevated

Answer 28

(B) Excited, ground An atom in the excited state will emit a photon to release energy and return to the ground state.

Question 29

Based on the Bohr model and the discrete levels of energy electrons can have, what is unique about each element based on electrons returning from the excited to the ground state? (A) Ground quantum number (B) Lyman Series (C) Angular momentum (D) Atomic emission spectrum

Answer 29

(D) Atomic emission spectrum Based on the Bohr model and the discrete levels of energy electrons can have, each element will have a unique atomic emission spectrum based on electrons returning from the excited to the ground state. This also means each element will have a unique absorption spectrum as well.