15.5 The Photoelectric Effect

1. ANo electrons leave, as nothing applies a potential difference to the plate
   A student who thinks a battery must pull the electrons out picks this. The energy for emission comes from the light itself; a potential difference in the laboratory apparatus only affects whether the emitted electrons reach a collecting plate.
2. BPositive charges are ejected, so the plate is left negatively charged
   A student who thinks positive charges are what move picks this. In a metal the positive nuclei stay in place; the photoelectric effect ejects electrons, so the plate becomes positively charged.
3. CElectrons leave the plate, which becomes positively charged Correct
   This is the photoelectric effect: incident radiation above the threshold frequency causes electrons to be emitted from the zinc. No circuit is needed for emission. The plate loses negative charge, so it is left positively charged.
4. DElectrons leave only after a delay, while energy builds up in the plate
   A student who thinks electrons gather energy from the light over time picks this. Each electron absorbs the energy of one photon; above the threshold frequency, emission begins without any delay.

CED 15.5.A.1 · Read this in Fix

1. ARed light of wavelength 650 nm from a bright lamp
   A student who thinks bright enough light always ejects electrons picks this. 650 nm is longer than 540 nm, so the frequency is below the threshold frequency and no electrons are ejected, however intense the light.
2. BOrange light of wavelength 600 nm left on for hours
   A student who thinks electrons can build up energy over time picks this. Each electron absorbs one photon at a time; photons of 600 nm light, below the threshold frequency, cannot eject electrons however long the light shines.
3. CInfrared radiation of wavelength 1200 nm from a lamp
   A student who thinks longer wavelengths carry more energy picks the longest wavelength. Longer wavelength means lower frequency; 1200 nm is far below the threshold frequency, so no electrons are ejected.
4. DViolet light of wavelength 410 nm from a very dim lamp Correct
   Frequency and wavelength are inversely related, f = c/λ, so light of wavelength shorter than 540 nm has a frequency above the threshold frequency. Violet light at 410 nm ejects electrons, however dim the lamp.

Working Threshold: f₀ = c/(540 nm). Light ejects electrons if f ≥ f₀, i.e. λ ≤ 540 nm, whatever its intensity or duration. Only 410 nm qualifies; 600 nm, 650 nm and 1200 nm are all longer than 540 nm.

CED 15.5.A.2 · Read this in Fix

1. AAn electron must gain at least 4.3 eV of energy to be removed from the metal Correct
   The work function is the minimum energy required to emit an electron from the metal. An electron needs at least 4.3 eV; a photon with more energy than that can eject one, and the surplus becomes kinetic energy.
2. BA photon of energy 5.0 eV cannot eject an electron from the metal
   A student who thinks the photon energy must match the work function exactly, as it must match an energy-level difference for absorption by an atom, picks this. Any photon with 4.3 eV or more can eject an electron; a 5.0 eV photon ejects one with up to 0.7 eV of kinetic energy.
3. CAll the electrons in the metal need the same energy, 4.3 eV, to be removed
   A student who reads the work function as the energy every electron needs picks this. It is the minimum: many electrons need more to escape, which is why emitted electrons have a range of kinetic energies.
4. DEach electron emitted from the metal leaves with 4.3 eV of kinetic energy
   A student who confuses the work function with the electrons' kinetic energy picks this. The work function is the energy used to free an electron; what is left of the photon's energy, at most hf − ϕ, is kinetic energy.

Working ϕ = 4.3 eV is the minimum energy to remove an electron: emission needs hf ≥ 4.3 eV, and emitted electrons have kinetic energies from 0 up to K_max = hf − 4.3 eV.

CED 15.5.A.3.i · Read this in Fix

1. AEmission stops once the light is dimmer than some minimum intensity
   A student who thinks there is a threshold intensity picks this. The threshold is a frequency. Above it, even very dim light ejects electrons; it simply ejects fewer each second.
2. BEmission continues without delay, but fewer electrons leave each second Correct
   Light at or above the threshold frequency induces emission regardless of the number of photons striking the plate. Dimmer light has fewer photons, so fewer electrons are emitted each second, but each photon can still free an electron at once.
3. CElectrons are emitted only after a delay that grows as the light dims
   A student who thinks electrons gather energy from the light over time picks this. Each electron absorbs a single photon, which has enough energy to free it, so there is no waiting, however dim the light.
4. DEmission continues without delay, but each electron has less kinetic energy
   A student who thinks the electrons' energy depends on the brightness of the light picks this. The frequency, and so the energy of each photon, is unchanged, so the electrons' kinetic energies are unchanged; only their number per second falls.

CED 15.5.A.2.i · Read this in Fix

1. AElectrons ejected by P and by Q have the same maximum kinetic energy Correct
   For both beams the current reaches zero at the same potential difference, −1.5 V, where even the fastest electrons are turned back. So K_max = e(1.5 V) = 1.5 eV for both. P gives the larger current because it delivers more photons each second: more electrons, but not faster ones.
2. BElectrons ejected by P have more maximum kinetic energy than Q's do
   A student who reads a larger current as more energetic electrons picks this. The current measures how many electrons arrive each second; the maximum kinetic energy is shown by where the current reaches zero, which is the same for both beams.
3. CBeam P has a higher frequency than beam Q, as it gives a larger current
   A student who thinks the current is set by the frequency picks this. Equal stopping potential differences mean equal K_max = hf − ϕ for the same metal, so the frequencies are equal; P's larger current means more photons per second.
4. DNeither beam ejects electrons from the plate while ΔV is below −1.5 V
   A student who equates zero current with zero emission picks this. Below −1.5 V the light still ejects electrons, but the potential difference turns every one of them back before it reaches the collecting plate.

CED 15.5.A.2.ii · Read this in Fix

1. AZ's electrons have the greatest K_max, and X's electrons the least
   A student who adds the work function to the photon energy picks this: 4.0 + 4.5 = 8.5 eV for Z. The work function is energy used up in freeing the electron, K_max = hf − ϕ, so a larger work function leaves less kinetic energy; Z emits nothing.
2. BX's electrons have the greatest K_max, and Z emits no electrons Correct
   K_max = hf − ϕ: for X, 4.0 eV − 2.0 eV = 2.0 eV; for Y, 4.0 eV − 3.0 eV = 1.0 eV. Z's work function is greater than the photon energy, so Z emits no electrons at all.
3. CAll three metals emit electrons, each with the same K_max
   A student who thinks each electron gets the photon's whole energy, 4.0 eV, picks this. At least the work function is used to free the electron, so the metals differ, and Z, with ϕ = 4.5 eV, emits no electrons.
4. DTheir K_max values cannot be ranked without the light's intensity
   A student who thinks the electrons' energy depends on the brightness of the light picks this. K_max = hf − ϕ depends only on the photon energy and the work function, both given.

Working K_max = hf − ϕ. X: 4.0 − 2.0 = 2.0 eV. Y: 4.0 − 3.0 = 1.0 eV. Z: 4.0 eV < 4.5 eV, so no emission. (hf + ϕ: X 6.0, Y 7.0, Z 8.5 eV. K = hf: all 4.0 eV.)

CED 15.5.A.3 · Read this in Fix

1. A5.0 eV
   A student who gives the electron the photon's whole energy picks this: hf = 5.0 eV. At least 2.3 eV is used to free the electron from the metal, so K_max = 5.0 eV − 2.3 eV = 2.7 eV.
2. B7.3 eV
   A student who adds the work function picks this: 5.0 eV + 2.3 eV. The work function is energy used up in freeing the electron, so it is subtracted: K_max = hf − ϕ.
3. C2.7 eV Correct
   The photon energy is hf = (4.14 × 10⁻¹⁵ eV·s)(1.2 × 10¹⁵ Hz) = 5.0 eV. K_max = hf − ϕ = 5.0 eV − 2.3 eV = 2.7 eV.
4. D2.3 eV
   A student who takes the work function to be the electrons' kinetic energy picks this. The work function is the energy needed to free an electron; the kinetic energy left over is hf − ϕ = 2.7 eV.

Working hf = (4.14 × 10⁻¹⁵ eV·s)(1.2 × 10¹⁵ Hz) = 4.97 eV. K_max = hf − ϕ = 4.97 eV − 2.3 eV = 2.67 eV ≈ 2.7 eV. (hf alone 5.0 eV; hf + ϕ = 7.3 eV; ϕ = 2.3 eV.)

CED 15.5.A.3.ii · Read this in Fix

1. A1.5
   A student who treats K_max as proportional to f picks this: 15.0/10.0 = 1.5. The line does not pass through the origin; K_max is proportional to f − f₀, which doubles.
2. B1.3
   A student who knows that at the threshold frequency the photon energy equals the work function, ϕ = hf₀, but adds the work function (K_max = hf + ϕ) picks this: (15.0 + 5.0)/(10.0 + 5.0) = 1.3. The work function is the energy used to free the electron, so it is subtracted: K_max = h(f − f₀), which doubles.
3. C1.0
   A student who thinks K_max depends on the intensity of the light, not its frequency, picks this, expecting no change. K_max = hf − ϕ rises with frequency; here it doubles.
4. D2.0 Correct
   The graph crosses the frequency axis at the threshold frequency, f₀ = 5.0 × 10¹⁴ Hz, and K_max = hf − ϕ = h(f − f₀). At 10.0 × 10¹⁴ Hz, f − f₀ = 5.0 × 10¹⁴ Hz; at 15.0 × 10¹⁴ Hz, f − f₀ = 10.0 × 10¹⁴ Hz. K_max doubles.

Working From the graph f₀ = 5.0 × 10¹⁴ Hz. K_max = h(f − f₀): K₂/K₁ = (15.0 − 5.0)/(10.0 − 5.0) = 2.0. (Proportional: 15.0/10.0 = 1.5; threshold ϕ = hf₀ with K_max = hf + ϕ (m10): 20.0/15.0 = 1.3; intensity unchanged: 1.0.)

CED 15.5.A.3.ii · Read this in Fix

1. A5.9 eV
   A student who uses −1.8 eV as K_max, keeping the sign read from the graph, picks this: ϕ = 4.1 eV − (−1.8 eV). A kinetic energy cannot be negative; K_max = e|ΔV_s| = 1.8 eV, so ϕ = 2.3 eV.
2. B1.8 eV
   A student who takes e times the stopping potential difference to be the work function picks this. eΔV_s = 1.8 eV is the maximum kinetic energy of the electrons; the work function is 4.1 eV − 1.8 eV.
3. C2.3 eV Correct
   The current reaches zero when the collecting plate is 1.8 V below the illuminated plate, so the fastest electrons have K_max = e(1.8 V) = 1.8 eV. The work function is ϕ = hf − K_max = 4.1 eV − 1.8 eV = 2.3 eV.
4. D4.1 eV
   A student who thinks a photon must have exactly the work function to eject an electron picks this. Photons with more energy than the work function also eject electrons, and the surplus, 1.8 eV here, is kinetic energy.

Working From the graph the current reaches zero at ΔV = −1.8 V, so ΔV_s = 1.8 V and K_max = eΔV_s = 1.8 eV. ϕ = hf − K_max = 4.1 eV − 1.8 eV = 2.3 eV. (Signed −1.8 eV: 5.9 eV; eΔV_s as ϕ: 1.8 eV; ϕ = hf: 4.1 eV.)

CED 15.5.A.3.iii · Read this in Fix

1. AElectrons are still ejected, but even the fastest are turned back Correct
   The light still ejects electrons; the collecting plate is at a lower potential, so the electrons slow down as they cross. At ΔV_s even the fastest just fail to reach the collector, so eΔV_s equals their kinetic energy, K_max.
2. BThe light has stopped ejecting any electrons from the illuminated plate
   A student who equates zero current with zero emission picks this. Emission depends only on the light and the metal; the potential difference only stops the emitted electrons from reaching the collecting plate.
3. CThe energy eΔV_s is equal to the work function of the metal
   A student who takes the stopping potential difference to measure the work function directly picks this. eΔV_s equals K_max; the work function is found from ϕ = hf − eΔV_s.
4. DEvery ejected electron has a kinetic energy of exactly eΔV_s
   A student who thinks all emitted electrons have the same kinetic energy picks this. The electrons leave with a range of kinetic energies; eΔV_s is the kinetic energy of the fastest, K_max.

CED 15.5.A.3.iii · Read this in Fix

1. Aϕ/e
   A student who thinks e times the stopping potential difference equals the work function picks this. eΔV_s equals the maximum kinetic energy of the emitted electrons, hc/λ − ϕ, not the energy needed to free them.
2. Bhc/(eλ)
   A student who thinks an emitted electron keeps the whole energy of the photon picks this, setting eΔV_s = hc/λ. At least ϕ is used to free the electron from the metal, so K_max = hc/λ − ϕ.
3. C(hc/λ−ϕ)/e Correct
   Each photon has energy hf = hc/λ, so the fastest electrons leave the plate with K_max = hc/λ − ϕ. The current just stops when the potential difference stops these electrons: eΔV_s = K_max, so ΔV_s = (hc/λ − ϕ)/e.
4. D(hc+ϕλ)/(eλ)
   A student who thinks the work function adds to the photon's energy, K_max = hf + ϕ, picks this: eΔV_s = hc/λ + ϕ. The work function is the energy needed to free the electron, so it is subtracted: K_max = hc/λ − ϕ.

Working Photon energy E = hf = hc/λ. K_max = hc/λ − ϕ. At the stopping potential difference the fastest electrons are just stopped: eΔV_s = K_max, so ΔV_s = (hc/λ − ϕ)/e. Errors: eΔV_s = ϕ → ϕ/e; K_max = hf with no work function → hc/(eλ); K_max = hf + ϕ → (hc+ϕλ)/(eλ).

CED 15.5.A.3.iii · Read this in Fix

1. A2hc/λ − K
   A student who takes the work function to be the kinetic energy of the emitted electrons sets ϕ = K and picks this. The work function is the energy needed to free an electron: ϕ = hc/λ − K, which gives K + hc/λ.
2. BK + hc/λ Correct
   From the first measurement, ϕ = hc/λ − K. Halving the wavelength doubles the photon energy to 2hc/λ, so the new K_max = 2hc/λ − ϕ = 2hc/λ − (hc/λ − K) = K + hc/λ.
3. C2K
   A student who takes K_max to be proportional to frequency doubles K because the frequency doubles. K_max = hf − ϕ: doubling f adds hf to K_max but does not double it, because ϕ is subtracted.
4. DK
   A student who thinks the electrons' kinetic energy is set by the intensity, which is unchanged, picks this. Each photon now carries twice the energy, 2hc/λ, so the fastest electrons leave with more kinetic energy.

Working Work function from the first measurement: K = hc/λ − ϕ, so ϕ = hc/λ − K. New photon energy: hc/(λ/2) = 2hc/λ. New K_max = 2hc/λ − ϕ = 2hc/λ − (hc/λ − K) = K + hc/λ. Errors: K_max ∝ f, f doubled → 2K; ϕ taken as K → 2hc/λ − K; K set by intensity, unchanged → K.

CED 15.5.A.3 · Read this in Fix

1. A6.2 eV
   A student who subtracts the wavelengths first, hc/(550 nm − 350 nm) = 1240/200 eV, picks this. Photon energy is inversely proportional to wavelength, so find each energy first: 3.54 eV − 2.25 eV = 1.3 eV.
2. B3.5 eV
   A student who gives the electron the photon's whole energy picks this: hc/λ = 3.54 eV. At least ϕ = hc/λ₀ = 2.25 eV is used to free the electron, leaving 1.3 eV.
3. C2.3 eV
   A student who takes the work function to be the kinetic energy of the emitted electrons picks this: hc/λ₀ = 2.25 eV. That is the energy needed to free an electron; the kinetic energy left over is 3.54 eV − 2.25 eV = 1.3 eV.
4. D1.3 eV Correct
   Light at the threshold wavelength has photons with just enough energy to free an electron, so ϕ = hc/λ₀ = (1240 eV·nm)/(550 nm) = 2.25 eV. Each 350 nm photon has hc/λ = 3.54 eV, so K_max = 3.54 eV − 2.25 eV = 1.3 eV.

Working At the threshold the photon energy equals the work function: ϕ = hc/λ₀ = (1240 eV·nm)/(550 nm) = 2.25 eV. Photon energy: hc/λ = (1240 eV·nm)/(350 nm) = 3.54 eV. K_max = 3.54 eV − 2.25 eV = 1.29 eV ≈ 1.3 eV. (hc/(λ₀ − λ) = 1240/200 = 6.2 eV; hc/λ alone 3.5 eV; ϕ taken as K_max 2.3 eV.)

CED 15.5.A.2 · Read this in Fix