15.2 The Bohr Model of Atomic Structure

1. AIt is positively charged and takes up most of the volume inside the atom.
   A student who takes diagrams of atoms at face value picks this. Diagrams enlarge the nucleus so that it can be seen; in a real atom the nucleus is tens of thousands of times smaller in radius than the atom.
2. BIt is neutral overall, since its neutrons cancel the charge of its protons.
   A student who thinks neutrons neutralize protons picks this. A neutron has no charge, so it cannot cancel anything: a nucleus with Z protons has charge +Ze, however many neutrons it has.
3. CIt is positively charged and takes up only a tiny fraction of the atom's volume. Correct
   The nucleus contains the atom's protons, so it is positively charged, and its radius is tens of thousands of times smaller than the atom's. Almost all of the atom's volume is the region around the nucleus where the electrons are found.
4. DIt keeps the electrons near it mainly through its gravitational pull on them.
   A student who pictures the atom as a small solar system picks this. The electrons are held by the electric force between the positive nucleus and the negative electrons; the gravitational force is about 10³⁹ times weaker.

CED 15.2.A.1.i · Read this in Fix

1. AAdding two protons to its nucleus
   A student who reads the upper numbers as the numbers of protons picks this: 16 − 14 = 2. The upper number is the mass number; the atomic numbers are the lower numbers, 7 and 8, so one extra proton is enough.
2. BAdding two neutrons to its nucleus
   A student who thinks the mass number identifies the element picks this, raising A from 14 to 16. The result, ¹⁶₇N, still has 7 protons, so it is an isotope of nitrogen, not oxygen.
3. CAdding one electron to the atom
   A student who thinks the number of electrons identifies the element picks this, since a neutral oxygen atom has 8 electrons. Adding an electron to nitrogen makes a negative ion of nitrogen: it still has 7 protons.
4. DAdding one proton to its nucleus Correct
   An element is identified by its number of protons. Nitrogen has 7 and oxygen has 8, so adding one proton makes the atom oxygen. The result has 8 protons and 7 neutrons: oxygen-15, an isotope of oxygen rather than oxygen-16 (and, with its 7 electrons, a positive ion of charge +e), but it is oxygen.

CED 15.2.A.2 · Read this in Fix

1. AThe accepted present-day picture of the atom, in which the electrons really move in circles
   A student who takes the planetary picture as literally true picks this. The Bohr model is a historical model; it is taught because it led to discrete energy states for hydrogen, not because electrons really follow circular tracks.
2. BA model in which gravity between the electron and the nucleus sets the size of each orbit
   A student who carries the solar-system analogy too far picks this. In the Bohr model the electron's orbit is determined by the electric force between the electron and the nucleus; gravity is far too weak to matter.
3. CA classical model in which the electron can orbit at any radius and so can have any energy
   A student who assumes that 'based on classical physics' means 'any orbit allowed' picks this. The importance of the Bohr model is that only certain orbits, and so only discrete energy states, are allowed.
4. DAn early model that used classical force and motion and led to discrete energy states for hydrogen Correct
   The Bohr model describes the electron's circular orbit with classical physics: the electric force and circular motion. It is a historical model, and it led to describing the hydrogen atom in terms of discrete energy states.

CED 15.2.A.3 · Read this in Fix

1. A34
   A student who thinks the electrons are inside the nucleus picks this, adding the 11 electrons to the 23 protons and neutrons. The electrons surround the nucleus; the nucleus holds only the 23 protons and neutrons.
2. B11
   A student who reads the lower number as the mass number picks this. In nuclear notation the lower number, 11, is the atomic number (protons); the upper number, 23, counts all the protons and neutrons in the nucleus.
3. C12
   A student who thinks the mass number also counts the electrons picks this: 23 − 11 electrons = 12. The mass number counts only protons and neutrons, so the nucleus has 23 particles; 12 is the number of neutrons alone.
4. D23 Correct
   The nucleus is made up of protons and neutrons, and the mass number, 23, is their total: 11 protons and 23 − 11 = 12 neutrons. The atom's 11 electrons are outside the nucleus.

Working The nucleus contains protons and neutrons only; their total is the mass number A = 23 (Z = 11 protons, A − Z = 12 neutrons). The 11 electrons are outside the nucleus.

CED 15.2.A.1.ii · Read this in Fix

1. A³⁷₁₇Cl Correct
   The atomic number Z is the number of protons, 17, written below. The mass number A is the number of protons plus neutrons, 17 + 20 = 37, written above. Electrons are not counted in either number.
2. B²⁰₁₇Cl
   A student who thinks the upper number counts neutrons picks this. The upper number is the mass number, which counts protons and neutrons together: 17 + 20 = 37.
3. C¹⁷₃₇Cl
   A student who has exchanged the two numbers picks this. The mass number, 37, goes above and the atomic number, 17, below; the mass number can never be smaller than the atomic number.
4. D⁵⁴₁₇Cl
   A student who counts the electrons in the mass number picks this: 17 + 20 + 17 = 54. The mass number counts only the particles in the nucleus, so it is 17 + 20 = 37.

Working Z = number of protons = 17. A = protons + neutrons = 17 + 20 = 37. Electrons (17) are not counted in A. Notation: ³⁷₁₇Cl.

CED 15.2.A.1.iii · Read this in Fix

1. ATwo protons were added to its nucleus.
   A student who thinks positive ions form by gaining protons picks this. Ions form when electrons are gained or lost; adding protons to the nucleus would make an atom of a different element.
2. BTwo electrons were removed from the atom. Correct
   Removing two electrons, each of charge −e, leaves the atom with two more protons than electrons, so its net charge is +2e. The nucleus is unchanged, so the atom is still the same element.
3. CTwo electrons were added to the atom.
   A student who links 'gain' with 'positive' picks this. Each electron carries charge −e, so adding two electrons gives a net charge of −2e, not +2e.
4. DTwo neutrons were removed from its nucleus.
   A student who thinks neutrons cancel the protons' charge picks this, reasoning that two protons are no longer cancelled. Neutrons have no charge, so removing them does not change the charge of the atom at all.

CED 15.2.A.1.iv · Read this in Fix

1. AThey interact in different ways, since the extra neutrons in chlorine-37 take part in bonding.
   A student who thinks any change in an atom's make-up changes its behavior picks this. The neutrons are inside the tiny nucleus; other atoms interact with the electrons, which are the same for both isotopes.
2. BChlorine-37 holds its electrons more tightly, since its more massive nucleus pulls harder.
   A student who thinks the nucleus holds its electrons by gravity picks this. The electrons are held by the electric force, which depends on the nuclear charge, +17e for both isotopes; the extra neutrons add mass but no charge.
3. CThey interact in the same way, since both have 17 electrons arranged in exactly the same way. Correct
   How an atom interacts with other atoms depends on its electrons. Both isotopes have 17 protons, so their neutral atoms have the same 17 electrons in the same arrangement, and they interact in the same way, apart from very small effects of their different masses.
4. DChlorine-37 cannot form bonds as chlorine-35 does, since as an isotope it is radioactive.
   A student who thinks 'isotope' means 'radioactive' picks this. Isotopes are simply atoms with the same protons and different neutrons; chlorine-37 is stable, and in any case radioactive atoms bond just as other atoms of the element do.

Working Both isotopes have Z = 17, so each neutral atom has 17 electrons in the same arrangement. Interactions with other atoms depend on the electrons, so the isotopes interact in the same way; they differ only in neutrons (18 and 20), which are inside the nucleus.

CED 15.2.A.2.i · Read this in Fix

1. AThey are atoms of different elements, since their mass numbers are different.
   A student who thinks the mass number identifies the element picks this. The mass numbers are 3 and 4, but the element is set by the number of protons, which is two in both nuclei.
2. BThey are isotopes of one element, since each nucleus contains two protons. Correct
   Each nucleus has two protons, so both atoms are the same element (helium). X has one neutron and Y has two, so their mass numbers are 3 and 4: they are isotopes of helium.
3. CThey are two ions of one element, since their nuclei hold different numbers of particles.
   A student who confuses isotopes with ions picks this. An ion has a net charge because its electrons differ in number from its protons, and the diagram says nothing about electrons. These nuclei differ only in their numbers of neutrons, which makes them isotopes.
4. DNucleus Y carries less charge than X, since its extra neutron cancels part of it.
   A student who thinks neutrons cancel the protons' charge picks this. Neutrons are uncharged: both nuclei have two protons, so both have charge +2e.

Working Count protons: X has 2, Y has 2, so the same element (Z = 2, helium). Neutrons: X has 1 (A = 3), Y has 2 (A = 4). Same Z, different A → isotopes. Nuclear charge +2e in both.

CED 15.2.A.2.ii · Read this in Fix

1. AThe claim is wrong: a proton or a neutron has nearly two thousand times an electron's mass. Correct
   A proton's mass (1.67 × 10⁻²⁷ kg) is about 1800 times an electron's (9.11 × 10⁻³¹ kg), and a neutron's is almost the same as a proton's. So the protons and neutrons in the tiny nucleus hold almost all of the atom's mass, even though the electrons occupy almost all of its volume.
2. BThe claim is wrong: electrons and nucleons have similar masses, so the mass is shared evenly.
   A student who thinks all three particles have similar masses picks this. An electron has about 1/1800 of a proton's mass, so the electrons contribute only a tiny fraction of the atom's mass; it is not shared evenly.
3. CThe claim is right: the part of the atom that fills nearly all its volume has most of the mass.
   A student who expects mass to follow size picks this. The electrons do occupy nearly all of the volume, but their mass is tiny; the small nucleus holds nearly all of the mass.
4. DThe claim is wrong: the nucleus fills most of the atom's volume, so most of the mass is in it.
   A student who pictures the nucleus as large picks this. The conclusion is right, but the reason is not: the nucleus is tens of thousands of times smaller than the atom in radius. It holds most of the mass because protons and neutrons are much more massive than electrons.

CED 15.2.A.2.iii · Read this in Fix

1. ADownward, toward the nucleus, and equal to the electric force exerted by the nucleus Correct
   The only force exerted on the electron in the Bohr model is the electric force from the positive nucleus, directed toward it, which is downward at the instant shown. That force is the net force: it provides the centripetal acceleration, F_e = mv²/r.
2. BDownward, toward the nucleus, and larger than the electric force, since F_c adds to it
   A student who treats the centripetal force as an extra force picks this. The centripetal force is not a second force; it is the name for the net force toward the center, which here is just the electric force.
3. CTo the right, along v, since a force is needed to keep the electron moving
   A student who thinks motion needs a force in its direction picks this. No object pushes the electron along its path; the only force is the electric pull toward the nucleus, perpendicular to v.
4. DZero, since an outward force on the electron balances the electric force
   A student who believes in an outward centrifugal force picks this. The electron's velocity is changing direction, so the net force cannot be zero; no object exerts an outward force on the electron.

CED 15.2.A.3.i · Read this in Fix

1. A3.1 × 10⁶ m/s
   A student who sets the kinetic energy equal to the magnitude of the electric potential energy, ½m_e v² = ke²/r, picks this; it is √2 times the correct speed. That is the condition for escaping, not for a circular orbit: the force condition gives ½m_e v² = ke²/(2r).
2. B5.1 × 10⁴ m/s
   A student who puts the proton's mass into m v²/r picks this. The mass in F_net = mv²/r is the mass of the object moving in the circle, the electron, whose mass is about 1800 times smaller.
3. C2.2 × 10⁶ m/s Correct
   The electric force exerted by the proton is the net force on the electron: ke²/r² = m_e v²/r, so v = √(ke²/(m_e r)) = √((9.0 × 10⁹)(1.60 × 10⁻¹⁹)²/((9.11 × 10⁻³¹)(5.3 × 10⁻¹¹))) = 2.2 × 10⁶ m/s.
4. D1.6 × 10¹ m/s
   A student who writes the electric force as ke²/r picks this: ke²/r = m_e v²/r gives v = √(ke²/m_e) ≈ 16, with no radius in it. The electric force varies as 1/r², and a correct check of units shows that ke²/m_e cannot give m²/s².

Working ke²/r² = m_e v²/r → v = √(ke²/(m_e r)) = √(9.0e9 × (1.60e-19)² / (9.11e-31 × 5.3e-11)) = √(4.77 × 10¹²) = 2.2 × 10⁶ m/s. (½m_e v² = ke²/r: 3.1 × 10⁶ m/s; m_p in place of m_e: 5.1 × 10⁴ m/s; F = ke²/r: 1.6 × 10¹ m/s.)

CED 15.2.A.3.i · Read this in Fix

1. A2.00
   A student who keeps the electric force the same in the new orbit picks this: with F fixed, v² = Fr/m is proportional to r, so v doubles. But F_e = ke²/r² is 1/16 as large at 4r₀, and v is halved.
2. B0.50 Correct
   The electric force provides the net force: ke²/r² = mv²/r, so v² = ke²/(mr) and v is proportional to 1/√r. Multiplying r by 4 divides v by √4 = 2, so v/v₀ = 0.50.
3. C4.00
   A student who treats the orbits like points on a turning wheel picks this, with v proportional to r. Separate orbits are not linked like a rigid wheel; the force condition gives v proportional to 1/√r.
4. D1.00
   A student who writes the electric force as ke²/r picks this: ke²/r = mv²/r gives v² = ke²/m, with no r left. The force varies as 1/r², so v depends on r as 1/√r.

Working ke²/r² = mv²/r → v² = ke²/(mr), v ∝ 1/√r. v/v₀ = √(r₀/(4r₀)) = 1/2 = 0.50.

CED 15.2.A.3.i · Read this in Fix

1. AF_NE > F_EN, since the nucleus has twice the electron's charge.
   A student who thinks the larger charge exerts the larger force picks this. The product of the two charges appears once in Coulomb's law, and it gives the magnitude of both forces of the pair.
2. BF_NE > F_EN, since the nucleus is far more massive than the electron.
   A student who thinks the more massive object exerts the larger force picks this. The forces are equal; the electron's much smaller mass means it has the much larger acceleration.
3. CF_EN > F_NE, since the electron moves and the nucleus stays nearly still.
   A student who thinks the moving object exerts the larger force picks this. Motion does not affect Newton's third law: the two forces of an interaction are always equal in magnitude.
4. DF_NE = F_EN, since they are the two forces of a single interaction. Correct
   The electron and the nucleus exert electric forces on each other that form a Newton's third law pair: equal in magnitude, opposite in direction. Both magnitudes equal k(2e)(e)/r².

Working Newton's third law: the forces of one interaction are equal in magnitude. F_NE = F_EN = k(2e)(e)/r². The electron's acceleration is about 7300 times the nucleus's.

CED 15.2.A.3.i · Read this in Fix

1. AThe circumference of the orbit is exactly five of the electron's de Broglie wavelengths. Correct
   The wave has five crests and five troughs around the orbit. One wavelength contains one crest and one trough, so five wavelengths fit around the circumference: 2πr = 5λ. An orbit is allowed only when a whole number of wavelengths fits in this way.
2. BThe circumference of the orbit is exactly ten of the electron's de Broglie wavelengths.
   A student who counts every crest and every trough as a whole wavelength picks this. Each crest or trough is half a wavelength, so the ten of them make five wavelengths.
3. CThe electron travels along the solid wavy line, moving in and out across the orbit.
   A student who reads the wave as a path picks this. The wave represents the electron's de Broglie wave, which shows how many wavelengths fit around the orbit; it is not a track the electron follows.
4. DAny other orbit radius would also be allowed, with the wave stretched to fit.
   A student who thinks any orbit is allowed picks this. A wave stretched to fit would not have a whole number of wavelengths matching the electron's de Broglie wavelength; only orbits with 2πr = nλ are allowed, which is why the energy states are discrete.

CED 15.2.A.3.ii · Read this in Fix

1. A4.00
   A student who makes the wavelength proportional to the radius alone picks this. The number of wavelengths that fit around the orbit also changes, from 1 to 2, so λ = 2πr/n grows by 4/2 = 2.
2. B0.25
   A student who thinks an electron in a larger orbit moves faster picks this: with v proportional to r, v would be 4 times as large and λ = h/(mv) 1/4 as large. The electron moves more slowly in the larger orbit, so its wavelength is longer.
3. C2.00 Correct
   The standing-wave condition is 2πr = nλ, so λ = 2πr/n. For n = 2 the radius is 4 times as large and n is 2 times as large, so λ₂/λ₁ = 4/2 = 2.00.
4. D1.00
   A student who thinks the de Broglie wavelength is a fixed property of the electron picks this. λ = h/p depends on the electron's momentum, which is different in each orbit.

Working 2πr = nλ → λ = 2πr/n. λ₂/λ₁ = (r₂/2)/(r₁/1) = (4r₁/2)/r₁ = 2.00. (Check with λ = h/(mv): v ∝ 1/√r, so v₂ = v₁/2 and λ₂ = 2λ₁.)

CED 15.2.A.3.ii · Read this in Fix

1. An²h²M/(4π²km²e²)
   A student who uses the proton's mass in mv²/r picks this, writing ke²/r² = Mv²/r while using the electron's momentum mv in its de Broglie wavelength. The orbiting object is the electron, so the mass in mv²/r is m, and the proton's mass does not appear.
2. Bnh/(2πe√(km))
   A student who writes the electric force as ke²/r picks this: ke²/r = mv²/r gives mv² = ke², and with v = nh/(2πmr) this gives r = nh/(2πe√(km)). The electric force varies as 1/r², so mv² = ke²/r.
3. Cn²h²/(4π²kme²) Correct
   The electric force is the net force: ke²/r² = mv²/r, so mv² = ke²/r. The standing-wave condition 2πr = nλ with λ = h/(mv) gives v = nh/(2πmr). Substituting, n²h²/(4π²mr²) = ke²/r, so r = n²h²/(4π²kme²): the allowed radii grow as n².
4. Dn²h²/(16π²kme²)
   A student who uses the condition for a string fixed at both ends, 2πr = nλ/2, picks this: v = nh/(4πmr), which gives r = n²h²/(16π²kme²). The orbit is a closed loop with no fixed ends, so its circumference holds n whole wavelengths: 2πr = nλ.

Working Force: ke²/r² = mv²/r, so mv² = ke²/r. Standing wave: 2πr = nλ = nh/(mv), so v = nh/(2πmr). Substituting: m·n²h²/(4π²m²r²) = ke²/r → n²h²/(4π²mr²) = ke²/r → r = n²h²/(4π²kme²) (5.30 × 10⁻¹¹ m for n = 1 with SI values). Errors: M in Mv²/r with the electron's momentum in λ → n²h²M/(4π²km²e²); F = ke²/r → mv² = ke² → r = nh/(2πe√(km)); string condition 2πr = nλ/2 → v = nh/(4πmr) → r = n²h²/(16π²kme²).

CED 15.2.A.3.ii · Read this in Fix

1. Ake²/(2r)
   A student who counts only the electron's kinetic energy picks this. ke²/(2r) is K, but the atom is a system of electron and proton, and its energy also includes their electric potential energy, −ke²/r.
2. B(3/2)ke²/r
   A student who puts only the magnitudes of the charges into U_E = kq₁q₂/r takes U_E = +ke²/r, adds K = ke²/(2r), and picks this. With q₁ = +e and q₂ = −e, U_E = −ke²/r, so the total is ke²/(2r) − ke²/r = −ke²/(2r).
3. C−ke²/(2r) Correct
   From F_e = mv²/r, mv² = ke²/r, so K = ke²/(2r). The electric potential energy is U_E = k(+e)(−e)/r = −ke²/r. The total, K + U_E, is −ke²/(2r): negative, as for a bound system, and half of U_E.
4. D−ke²/r
   A student who takes the atom's energy to be the electric potential energy alone picks this. The electron is moving, so the system's energy also includes its kinetic energy, ke²/(2r), and the total is −ke²/(2r).

Working Force: ke²/r² = mv²/r, so mv² = ke²/r and K = ½mv² = ke²/(2r). Electric potential energy: U_E = k(+e)(−e)/r = −ke²/r. Total: E = K + U_E = ke²/(2r) − ke²/r = −ke²/(2r). Errors: kinetic energy alone → ke²/(2r); electric potential energy alone → −ke²/r; U_E taken as +ke²/r (magnitudes of the charges) → E = ke²/(2r) + ke²/r = (3/2)ke²/r.

CED 15.2.A.3 · Read this in Fix

1. Ah/√(2kme²/r)
   A student who finds the speed by setting ½mv² = ke²/r picks this; that speed, and so the momentum, is √2 times too large, and the wavelength √2 times too short. The circular orbit is set by the force condition ke²/r² = mv²/r.
2. B2hr/(ke²)
   A student who takes the wavelength to be h divided by the kinetic energy, ke²/(2r), picks this. The p in λ = h/p is the momentum mv; h/K has units of time, not length.
3. C2hcr/(ke²)
   A student who uses the photon relationship λ = hc/E with the electron's kinetic energy, ke²/(2r), picks this. That relationship holds for photons, which travel at c; the electron's wavelength is h/(mv).
4. Dh/√(kme²/r) Correct
   The electric force is the net force: ke²/r² = mv²/r, so mv² = ke²/r and the momentum is p = mv = √(kme²/r). Then λ = h/p = h/√(kme²/r): a larger orbit means a smaller momentum and a longer wavelength.

Working Force: ke²/r² = mv²/r, so mv² = ke²/r. Momentum: p = mv = √(m·mv²) = √(kme²/r). de Broglie: λ = h/p = h/√(kme²/r). Errors: ½mv² = ke²/r → p = √(2kme²/r) → h/√(2kme²/r); λ = h/K with K = ke²/(2r) → 2hr/(ke²), which has units of time; photon relationship λ = hc/K → 2hcr/(ke²).

CED 15.2.A.3.i · Read this in Fix

1. A3.64 × 10⁵ m/s
   A student who uses the string condition, 2πr = nλ/2, picks this; it doubles λ and so halves v. The orbit is a closed loop, so its circumference holds whole wavelengths: 2πr = nλ.
2. B2.43 × 10⁵ m/s
   A student who ties the wavelength to the radius alone, fitting one wavelength around this orbit as in the smallest orbit, picks this: λ = 2πr. In the n = 3 state three wavelengths fit around the orbit, so λ = 2πr/3.
3. C7.28 × 10⁵ m/s Correct
   Three whole wavelengths fit around the orbit: λ = 2πr/3 = 9.99 × 10⁻¹⁰ m. Then v = h/(m_eλ) = 7.28 × 10⁵ m/s.
4. D1.21 × 10³ m/s
   A student who takes λ = h/K picks this, finding K = h/λ and then v = √(2K/m_e). The de Broglie wavelength is h divided by the momentum, so v = h/(m_eλ).

Working Standing wave: 2πr = nλ, so λ = 2πr/3 = 2π(4.77 × 10⁻¹⁰ m)/3 = 9.99 × 10⁻¹⁰ m. de Broglie: v = h/(m_eλ) = (6.63 × 10⁻³⁴ J·s)/((9.11 × 10⁻³¹ kg)(9.99 × 10⁻¹⁰ m)) = 7.28 × 10⁵ m/s. (The force model, v = √(ke²/(m_e r)), gives the same speed.) Errors: string condition 2πr = nλ/2 → 3.64 × 10⁵ m/s; one wavelength around the orbit, λ = 2πr → 2.43 × 10⁵ m/s; λ = h/K, so K = h/λ and v = √(2K/m_e) → 1.21 × 10³ m/s.

CED 15.2.A.3.ii · Read this in Fix