Physics



AP Physics 6: Subatomic Mass, Energy & Momentum Name __________________________

A. Atomic Nucleus

1. 2 types of particles—nucleons

a. positively charged proton and neutral neutron

b. rest mass chart (1 u = 1.66 x 10-27 kg)

|Object |kg |u |

|Proton (11p) |1.67262 x 10-27 |1.007276 |

|Neutron (10n) |1.67493 x 10-27 |1.008665 |

|Electron (0-1e) |9.1094 x 10-31 |0.00054858 |

2. nuclide (combination of protons and neutrons)

a. number of protons = atomic number (Z)

b. number of protons + neutrons = mass number (A)

c. nuclide symbol: AZX

1. X is atomic symbol with same Z number

2. A varies = isotopes

3. binding energy (BE) and nuclear forces

a. takes energy to break a stable nuclide into its parts ∴ energy is added to the nuclide to separate into protons and neutrons

1. nuclide + BE = protons + neutrons

2. change in mass, E = mc2

a. c = 3 x 108 m/s (speed of light)

b. if E = BE, then m = mBE in kg

c. mnuclide + mBE = mp + mn

b. binding energy per nucleon (BE/# nucleons)

[pic]

c. nuclear processes

1. exothermic when BE/#nucleon increases

2. mproducts – mreactants = mBE (-mBE is exothermic)

3. nuclear fusion: Zsmall → Zlarge

a. thermonuclear devise—Hydrogen bomb

b. sun's energy

4. nuclear fission: Zlarge → Zsmall

a. atomic bomb (Hiroshima)

b. nuclear power plants

4. strong nuclear force

a. acts over a very short distance (10-18 m)

b. bind "up" and "down" quarks together

1. proton = 2 up (u+2/3) + 1 down (d-1/3) quark

2. neutron = 1 up (u+2/3) + 2 down (d-1/3) quark

proton neutron

c. binds neutrons and protons together

d. neutrons stabilize nuclide

1. electric repulsion between protons is significant when protons > 10-18 m apart

2. neutrons increase total strong nuclear force without increasing electric repulsion

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5. naturally occurring radioactivity

a. unstable nuclides undergo nuclear change

b. nuclide is too large (Z > 84)

1. alpha radiation, 42α (42He)—low penetration

2. 22688Ra → 22286Rn + 42α

3. BE/# nucleons increases

c. A > average atomic mass

1. beta radiation, 0-1β (0-1e)—medium penetration

a. -⅓d → ⅔u + 0-1β

b. 10n → 11p + 0-1β-

c. involves "weak nuclear force" and generates an additional particle called neutrino (ν)

2. 146C → 147N + 0-1β

d. A < average atomic mass

1. positron radiation, 01β

a. ⅔u → -⅓d + 01β

b. 11p → 10n + 01β

2. 116C → 115B + 01β

e. nuclide falls from high energy to low energy state

1. gamma radiation, 00γ—high penetration

2. matter-antimatter collision

a. 0-1e + 01e → γ

b. E = 2mec2

6. artificial induced radioactivity (transmutation)

a. nuclide bombarded by subatomic particle (proton, neutron, alpha particle, etc.)

b. nuclear reaction

1. mass and charge are conserved

2. 10n + 147N → 146C + 11p

7. rate of decay and half-life

a. time it takes to reduce radioactivity by half is constant = half life, t½

b. 1 → 1/2 → 1/4 → 1/8 → 1/16 ...

B. Photons and Electrons

1. photon

a. quantum unit of electromagnetic energy

b. wave property, c = λf

1. wavelength, λ (m)

2. frequency, f (s-1)

c. energy, E = hf

[pic]

1. Planck's constant, h = 6.63 x 10-34 J•s

2. electromagnetic radiation

a. gamma > x-rays > UV > visible > IR > radio

b. violet > blue > green > yellow > orange > red

3. brightness measures intensity (not energy)

4. warm objects glow infrared (hot = visible light)

d. relativistic mass, m = E/c2 (mo = 0)

e. momentum, p = mc

f. interconnecting formulas (kg, m, s, J)

|In terms of: |m |f |λ |E/p |

|Energy |E = mc2 |E = hf |E = hc/λ |E = pc |

|Momentum |p = mc |p = hf/c |p = h/λ |p = E/c |

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2. electron

a. electron properties

1. same charge as proton, but negative

2. mo = 9.11 x 10-31 kg (0.00054858 u)

b. electron structure in an atom: Bohr model

[pic]

1. electrons occupy discrete energy levels

a. En = -B/n2 (BH = 13.6 eV)

b. n = energy level #

c. electron volt (eV) = small unit of energy

1 eV = 1.6 x 10-19 J

2. electrons change n by absorbing/emitting photons

a. Ephoton = En-high – En-low

b. EeV = 1240 eV•nm/λnm

c. photoelectrons

1. Ephoton > ionization energy ∴ atom emits electron

2. Einstein's photoelectric effect equation

[pic]

a. Kelectron = Ephoton – φ

|Kelectr| | |

|on | | |

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| |threshold fphoton |

b. φ called "work function" = ionization energy

1. minimum energy for photoelectrons

2. threshold λnm = 1240 eV•nm/φeV

c. electron can be stopped by a reverse voltage

1. called stop voltage, Vstop

2. Vstop = Kelectron (in eV)

3. (KeV)(1.6 x 10-19) = KJ

KJ = ½(9.11 x 10-31 kg)v2

d. De Broglie wavelength: λparticle = h/p = h/mov

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A. Atomic Nucleus

Questions 1-21 Briefly explain your answer.

1. There are 82 protons in a lead nucleus. Why doesn't the lead nucleus burst apart?

(A) electric repulsive forces don't exist in the nucleus

(B) gravity overpowers the electric repulsion

(C) neutrons neutralize the positively charged protons

(D) none of the above

|D |Strong nuclear forces > electric repulsion |

2. What weighs more, A—an electron and a proton, or B—a hydrogen atom?

(A) A (B) B (C) tie

|A |It takes energy to remove an electron from neutral atom ∴ separate parts |

| |have more total mass |

Questions 3-5 radiation passes through a vacuum, where one side is positively charged and the other is negatively charged. ++++++++++++++++++++

A

B

_ _ _ _ _ _ _ _ _ _ _ _ _ _ C

3. Which is the path of an alpha particle?

|C |Negative charge attracts + alpha particle |

4. Which is the path of a beta particle?

|A |Positive charge attracts – beta particle |

5. Which is the path of gamma radiation?

|B |Gamma is uncharged ∴ not attracted to + or – charge |

Questions 6-9 Refer to the nonspontaneous nuclear decay:

23892U → 23490Th + 42He

6. Which weighs more, A—U-238, or B—Th-234 + He-4?

(A) A (B) B (C) tie

|B |Nonspontaneous ∴ energy (mass) is added to reactants form products |

7. Which has greater momentum, A—Th-234, or B—He-4?

(A) A (B) B (C) tie

|C |Equal impulse (FΔt) at separation ∴ equal momentum |

8. Which has greater velocity, A—Th-234, or B—He-4?

(A) A (B) B (C) tie

|B |Same momentum, but He has less mass ∴ greater velocity |

9. Which has greater kinetic energy, A—Th-234, or B—He-4?

(A) A (B) B (C) tie

|B |mHevHe = mThvTh, but in kinetic energy, v is squared ∴ He has more kinetic|

| |energy |

10. What is produced by the beta decay of 31H?

(A) 21H (B) 11H (C) 32He (D) 42He

|C |31H → 0-1β + 32He |

11. What is produced by the alpha decay of 21084Po?

(A) 21082Pb (B) 20682Pb (C) 21086Rn (D) 21486Rn

|B |21084Po → 42α + 20682Pb |

12. Complete the nuclear equation: 146C + 0-1e → ___

(A) 156C (B) 157N (C) 145B (D) 147N

|C |146C + 0-1e → 145B |

13. Complete the nuclear equation: 10n + 168O → ___ + 21H

(A) 178O (B) 158O (C) 157N (D) 159F

|C |10n + 168O → 157N + 21H |

Questions 14-15 16 g of radioactive material has a 30-yr half-life.

14. How much is left after 90 years?

(A) 8 g (B) 4 g (C) 2 g (D) 1 g

|C |90 years = 3 t½ (16 g → 8 g → 4 g → 2 g) |

15. How long will it take to reduce the amount to 8 g?

(A) 30 yr (B) 60 yr (C) 90 yr (D) 120 yr

|A |16 g → 8 g = 1 t½ (30 yr) |

16. What is the half life of a radioactive material that decays to ¼ the original amount in 16 years?

(A) 1 yr (B) 2 yr (C) 4 yr (D) 8 yr

|D |1 → ½ → ¼ (2 t½) ∴ 16 yr/2 = 8 yr |

17. Which is more radioactive, substances A (t½ = 100 s) or B (t½ = 50 s)?

(A) A (B) B (C) tie

|B |Shorter t½ equals more radioactive |

18. Which type of radiation goes farther in matter before losing all of its energy?

(A) alpha (B) beta (C) gamma

|C |Gamma radiation is pure energy ∴ with charge or mass, it can penetrate |

| |matter without much loss in energy |

19. Define

|nucleon |Subatomic particle found in the nucleus |

|nuclide |Combination of protons and neutrons |

|Z # |Atomic number = number of protons |

|A # |Mass number = number of protons + neutrons |

|isotope |same Z value, but different A value |

20. Consider a carbon-12 atom (a.m. = 12.0 u).

a. What is the mass in kg?

|12.0 u x 1.66 x 10-27 kg/1 u = 1.99 x 10-26 kg |

b. What is the energy equivalence in J?

|E = mc2 = (1.99 x 10-26 kg)(3 x 108 m/s)2 = 1.79 x 10-9 J |

21. Complete the following chart for the isotope, U-235.

|Number of |Number of |Number of |Z value |A value |

|nucleons |protons |neutrons | | |

|235 |92 |143 |92 |235 |

22. Consider the helium nuclide (42He = 4.001504). Determine

a. the difference in mass between nucleons and nuclide.

(1) in u

|Δm = 2(mp) + 2(mn) – mHe |

|Δm = 2(1.007276) + 2(1.008665) – 4.001504 = 0.030378 u |

(2) in kg

|0.030378 u x 1.66 x 10-27 kg/1u = 5.04 x 10-29 kg |

b. the total binding energy in J.

|E = Δmc2 = (5.04 x 10-29 kg)(3 x 108 m/s)2 = 4.54 x 10-12 J |

c. the binding energy per nucleon.

|4.54 x 10-12 J/4 = 1.13 x 10-12 J |

23. Complete the following chart.

| |Symbol |Mass |Charge |Penetrating ability |

|alpha |42He |4 |2 |low |

|beta |0-1e |0 |-1 |medium |

|gamma |γ |0 |0 |high |

24. Fill in the missing symbols.

|10B + n → α + 7Li |1H + n → γ + 2H |50Cr + n → γ + 51Cr |

25. Consider the alpha decay of U-238: 238U → 4He + 234Th

|238U = 238.050784 u |4He = 4.002602 u |234Th = 234.055381 u |

a. Determine the change in mass per U-238 atom.

(1) in u

|Δm = mHe + mTh - mU |

|Δm = 4.002602 u+234.055381 u–238.050784 u=0.007199 u |

(2) in kg

|0.007199 u x 1.66 x 10-27 kg/1 u = 1.20 x 10-29 kg |

b. Is the reaction spontaneous?

|Δm > 0 ∴ reaction is not spontaneous. |

c. Determine the change in energy (in J) per atom.

|E = mc2 |

|E = (1.20 x 10-29 kg)(3 x 108 m/s)2 = 1.08 x 10-12 J |

26. A sample of 100Pd (half-life 4 days) has mass 1.776 g at 3:00 P.M. on July 4; what mass of 100Pd remains at 3:00 P.M. on July 16?

|12 days = 3 t½ |

|(1/8)1.776 g = 0.222 g |

27. Consider a helium-4 atom (atomic mass = 4.00 u).

a. What is the mass in kg?

|4.00 u x 1.66 x 10-27 kg/1 u = 6.64 x 10-27 kg |

b. What is the energy equivalence in J?

|E = mc2 |

|E = (6.64 x 10-27 kg)(3 x 108 m/s)2 = 5.98 x 10-10 J |

28. Complete the following chart for the isotope, Rn-226.

|Number of |Number of |Number of |Z value |A value |

|nucleons |protons |neutrons | | |

|226 |86 |140 |86 |226 |

29. Determine the binding energy (in joules) of an average nucleon in 3H (3.015500 u) by completing the following sequence of calculations

|Δm |Δm = 1mp + 2mn – mH-3 |

|(u) |Δm = (1.007276) + 2(1.008665) – (3.015500) |

| |Δm = 0.009106 u |

|Δm |Δm = 0.009106 u x 1.66 x 10-27 kg/1 u |

|(kg) |Δm = 1.51 x 10-29 kg |

|BE |BE = mc2 = (1.51 x 10-29 kg)(3 x 108 m/s)2 |

| |BE = 1.36 x 10-12 J |

|BE/A |BE/A = 1.36 x 10-12 J/3 = 4.53 x 10-13 J/nucleon |

30. Fill in the missing symbols.

|27Al + α→ 1n + 30P |9Be + α → n + 12C |

31. Stationary Radium-226 (225.97709 u) undergoes an alpha decay (4.001504 u) to produce radon-222 (221.97036).

a. Write a nuclear equation for this radioactive decay.

|22688Ra → 22286Rn + 42He |

b. Calculate the change in mass in kg for this reaction.

|221.97036 + 4.001504 – 225.97709 = -0.005226 u |

|-0.005226 u x 1.66 x 10-27 kg/1 u = -8.675 x 10-30 kg |

c. Calculate the energy is joules released per decay.

|E = mc2 |

|E = (8.678 x 10-30 kg)(3 x 108 m/s)2 = 7.81 x 10-13 J |

d. If all the released energy becomes kinetic energy, what is Kα/KRa?

|mRavRa = mαvα → (222)(4) = (4)(222) |

|Kα = ½mαvα2 = (4)(222)2 = 222 = 55.5 |

|KRa ½mRavRa2 (222)(4)2 4 |

e. What is the alpha particle's velocity if 98 % of the energy is absorbed by the particle?

|E = ½mv2 |

|(.98)(7.81 x 10-13 J) = ½(4 x 1.67 x 10-27 kg)v2 |

|v = 1.51 x 107 m/s |

f. What is the De Broglie wavelength of the alpha particle?

|λ = h/mv |

|λ = 6.63 x 10-34 J•s/(4 x 1.67 x 10-27 kg)(1.51 x 107 m/s) |

|λ = 6.57 x 10-15 m |

32. How long does it take for a sample of 32P (t1/2 = 14 days) to lose 7/8 of its activity?

|The loss of 7/8 activity means that 1/8 remains |

|It takes 3 half-lives to reach 1/8 (1 → 1/2 → 1/4 → 1/8) |

|3 x 14 = 42 days |

B. Photons and Electrons

33. Check which color has the greatest value for the following.

| |red |yellow |violet |can't tell |

|frequency | | | | |

|wavelength | | | | |

|energy | | | | |

|relativistic mass | | | | |

|momentum | | | | |

|intensity | | | | |

34. Complete the chart for an X-ray photon (λ = 1.54 x 10-10 m).

|f |c = fλ |

| |3 x 108 m/s = f(1.54 x 10-10 m) ∴ f = 1.95 x 1018 s-1 |

|E |E = hf |

| |E = (6.63 x 10-34 J•s)(1.95 x 1018 s-1) = 1.29 x 10-15 J |

|m |E = mc2 |

| |1.29 x 10-15 J = m(3 x 108 m/s)2 ∴ m = 1.44 x 10-32 kg |

|p |p = h/λ |

| |p = (6.63 x 10-34 J•s)/(1.54 x 10-10 m) = 4.31 x 10-24 kg•m/s |

35. A proton and antiproton (m = 1.67 x 10-27 kg) collide and convert all mass into photon energy. Determine the photon's

|E |E = mc2 = 2(1.67 x 10-27 kg)(3.0 x 108 m/s)2 |

| |E = 3.0 x 10-10 J |

|f |E = hf → f = E/h |

| |f = 3.0 x 10-10 J/6.63 x 10-34 J•s = 4.5 x 1023 s-1 |

|λ |c = fλ → λ = c/f |

| |λ = 3 x 108 m/s/4.5 x 1023 s-1 = 6.7 x 10-16 m |

|p |p = h/λ = (6.63 x 10-34 J•s)/(6.7 x 10-16 m) |

| |p = 1.0 x 10-18 kg•m/s |

Questions 36-47 Briefly explain your answer.

36. What is the ionization energy of ground state hydrogen?

(A) 0 eV (B) 13.6 eV (C) 41.2 eV (D) 54.4 eV

|B |E∝ – E1 = 0 – (-13.6 eV/12) = 13.6 eV |

37. Which transition results in the greatest gain in energy?

(A) 2 → 5 (B) 5 → 2 (C) 3 → 10 (D) 1 → 2

|D |The energy gap between 1 and 2 is ¾ of the total energy between ground |

| |state and ionization |

38. The Balmer series includes two blue, a blue-green and red spectral line. Which transition generates the red?

(A) 3 → 2 (B) 4 → 2 (C) 5 → 2 (D) 6 → 2

|A |Red would have the least energy ∴ the electron would come from the energy |

| |level closest to 2 |

39. A hydrogen electron is excited to n = 4. How many different transitions are possible to reach ground state?

(A) 1 (B) 2 (C) 3 (D) 6

|D |4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 → 1 |

Question 40-42 Metal A has a greater work function φ compared to metal B.

40. Which metal has the greater threshold wavelength?

(A) A (B) B (C) tie

|B |Greater wavelength equals less energy (E = hc/λ) ∴ B |

41. Which metal has the greater threshold frequency?

(A) A (B) B (C) tie

|A |Greater frequency equals more energy (E = hf) ∴ A |

42. Green light does NOT generate photoelectrons in metal A. Which color might be able to generate photoelectrons?

(A) red (B) orange (C) yellow (D) blue

|D |red < orange < yellow < green < blue < violet ∴ blue |

Questions 43-44 A metal surface emits photoelectrons when exposed to 400-nm light.

43. What happens if the metal is exposed to 300-nm light?

(A) more electrons are emitted

(B) fewer electrons are emitted

(C) more energetic electrons are emitted

(D) no electrons are emitted

|C |300-nm light has more energy than 400-nm light; the extra energy becomes |

| |electron kinetic energy |

44. What will happen if the metal is exposed to brighter 400-nm light?

(A) more electrons are emitted

(B) fewer electrons are emitted

(C) more energetic electrons are emitted

(D) no electrons are emitted

|A |Brighter, more intense light, has more photons, but not more energetic |

| |photons |

45. The speed of proton A is faster than proton B. Which one has the longer De Broglie wavelength?

(A) A (B) B (C) tie

|B |λ = h/mv → λ ∝ 1/v ∴ slower = longer wavelength |

46. A proton and electron have the same speed, which has the longer De Broglie wavelength?

(A) proton (B) electron (C) tie

|B |λ = h/mv → λ ∝ 1/m ∴ lighter = longer wavelength |

47. A proton and electron have the same momentum, which has the longer De Broglie wavelength?

(A) proton (B) electron (C) tie

|C |λ = h/p → λ ∝ 1/p∴ same p = same wavelength |

48. Red light has a wavelength of 750 nm. Determine the

|f |c = fλ |

| |3 x 108 m/s = f(750 x 10-9 m) ∴ f = 4.0 x 1014 s-1 |

|E |E = hf |

| |E = (6.63 x 10-34 J•s)(4.0 x 1014 s-1) = 2.7 x 10-19 J |

|m |E = mc2 |

| |2.7 x 10-19 J = m(3 x 108 m/s)2 ∴ m = 2.9 x 10-36 kg |

|p |p = h/λ |

| |p = (6.63 x 10-34 J•s)/(750 x 10-9 m) = 8.8 x 10-28 kg•m/s |

49. Complete the chart for a hydrogen transition from n = 2 → 3.

|E2 |E2 = -B/n2 = -13.6 eV/22 = -3.4 eV |

|E3 |E3 = -B/n2 = -13.6 eV/32 = -1.5 eV |

|ΔE |ΔE = -1.5 eV – (-3.4 eV) = 1.9 eV |

|λphoton |Ephoton = (1240 eV•nm)/λnm |

| |1.9 eV = (1240 eV•nm)/λnm ∴ λnm = 653 nm |

50. The hypothetical atom has four energy states as shown.

a. Calculate the change in energy for all transitions from the forth energy level to ground state.

|Energy|-4 | |n = 4 |Transition |ΔE |

|(eV) | | | | | |

| | | | |n = 4 → n = 3 |-16 - -4 = -12 eV |

| |-16 | |n = 3 | | |

| | | | |n = 4 → n = 2 |-36 - -4 = -32 eV |

| | | | | | |

| |-36 | |n = 2 |n = 4 → n = 1 |-144 - -4 = -140 eV |

| | | | | | |

| | | | |n = 3 → n = 2 |-36 - -16 = -20 eV |

| | | | | | |

| | | | |n = 3 → n = 1 |-144 - -16 = -128 eV |

| | | | | | |

| |-144 | |n = 1 |n = 2 → n = 1 |-144 - -36 = -108 eV |

| | | | | | |

b. Ground state atoms are irradiated with photons having energies between 120 eV and 130 eV. What energies can be emitted by atoms of this gas?

|20 eV, 128 eV, 108 eV |

51. Consider an electron transition from the second energy level to ground state on a hydrogen atom.

a. What are E1 and E2?

|E1 = -13.6 eV/12 = -13.6 eV |

|E2 = -13.6 eV/22 = -3.4 eV |

b. What is ΔE of the emission line?

|ΔE = En-high – En-low |

|ΔE = -3.4 – (-13.6) = 10.2 eV |

c. What is the wavelength of the photon?

|Ephoton = 1240/λnm |

|λnm = 1240/10.2 = 122 nm |

d. How much energy is needed to ionize an electron in the n = 2 state?

|ΔE = E∞ – En |

|ΔE = 0 – (-B)/n2 = 13.6 eV/22 = 3.4 eV |

52. What is the threshold wavelength that will emit electrons from a metal whose work function is 3.10 eV?

|λnm = 1240 eV•nm /φeV = 1240 eV•nm /3.10 eV = 400 nm |

53. A metal surface (φ = 2.40 eV) is exposed to 240 nm light.

a. What is the energy of the photons?

|E = 1240/λnm = 1240 eV•nm/240 nm = 5.17 eV |

b. What is the maximum kinetic energy (in eV) of electrons ejected from a surface?

|K = Ephoton - φ = 5.17 eV – 2.40 eV = 2.77 eV |

54. When 230-nm light falls on a metal, the photoelectrons are stopped with a voltage of 1.64 V.

a. What is the kinetic energy of the photoelectrons?

|Vstop = Kelectron = 1.64 eV |

b. What is the energy of the photons?

|Ephoton = 1240/λnm = 1240 eV•nm/230 nm = 5.39 eV |

c. What is the work function of the metal?

|φ = Ephoton – K = 5.39 – 1.64 = 3.75 eV |

55. A sodium surface (φ = 2.28 eV) is illuminated by 500 nm light.

a. What is the energy per photon?

|E = 1240 eV•nm/500 nm = 2.48 eV |

b. What is the maximum kinetic energy of the photoelectrons?

|K = Ephoton - φ = 2.48 eV – 2.28 eV = 0.20 eV |

c. What is the threshold wavelength for photoemission from the surface of sodium?

|φ = 1240 eV•nm/λnm = 1240 eV•nm/2.28 nm = 544 nm |

d What is the kinetic energy of the photoelectrons when 600 nm light shines on the sodium surface?

|0 eV, the energy of the photon is less than φ |

56. Determine the following with a wavelength of 500 nm.

| |photon |electron |

|λ |500 nm x 10-9 m/1 nm = 500 x 10-9 m |

|(m) | |

|p |p = h/λ |p = h/λ |

|(kg•m/s) |p = 6.63E-34/500E-9 |p = 6.63E-34/500E-9 |

| |p = 1.3E-27kg•m/s |p = 1.3E-27 kg•m/s |

|m |m = p/c |9.1E-31 kg |

|(Kg) |m = 1.3E-27/3E8 | |

| |m = 4.4E-36 kg | |

|v |3E8 m/s |v = p/m |

|(m/s) | |v = 1.3E-27/9.1E-31 |

| | |v = 1.4E3 m/s |

|f |f = c/λ |f = v/λ |

|(s-1) |f = 3E8/500E-9 |f = 1.4E3/500E-9 |

| |f = 6E14 |f = 2.8E9 |

|E |E = hf |E = ½mv2 |

|(J) |E = (6.63E-34)(6E14) |E = ½(9.1E-31)(1.4E3)2 |

| |E = 4E-19 J |E = 8.9E-25 J |

|E |4E-19 J x 1 eV |8.9E-25 J x 1 eV |

|(eV) |1.6E-19 J |1.6E-19 J |

| |E = 2.5 eV |E = 5.6E-6 eV |

Practice Multiple Choice (No calculator)

Briefly explain your answer.

1. The nuclear reaction X → Y + Z occurs spontaneously. If Mx, My, and Mz are the masses of the three particles, which of the following relationships is true?

(A) MX < MY – MZ (B) MX < MY + MZ

(C) MX > MY + MZ (D) MX – MY < MZ

|C |Spontaneous process loses energy (mass) in order to reach a lower energy |

| |state ∴ MX > MY + MZ |

2. A negative beta particle is emitted during the radioactive decay of 21482Pb. Which is the resulting nucleus?

(A) 21080Hg (B) 21481Tl (C) 21383Bi (D) 21483Bi

|D |21482Pb → 0-1β + xyZ |

| |214 = 0 + x ∴ x = 214, 82 = -1 + y ∴ y = 83 ∴ 21483Bi |

3. The deuteron (21H) mass md is related to the neutron mass mn and the proton mass mp by which of the following?

(A) md = mn + mp (B) md = mn + mp + mBE

(C) md = 2(mp) (D) md = mn + mp – mBE

|D |Taking a deuteron apart takes energy ∴ mBE, is added to the reactant side.|

| |md + mBE = mp + mn |

4. At noon the decay rate is 4,000 counts/minute. At 12:30 P.M. the decay rate is 2,000 counts/minute. The predicted decay rate in counts/minute at 1:30 P.M. is

(A) 0 (B) 500 (C) 667 (D) 1,000

|B |The counts/minute decrease by half in 30 min (t½) ∴ at 1:00 the count is |

| |1,000 and at 1:30 the count is 500 |

5. In the photoelectric effect, the maximum speed of electrons emitted by a metal surface when it is illuminated by light depends on which of the following?

I. Intensity of the light

II. Frequency of the light

III. Nature of the photoelectric surface

(A) I only (B) II only (C) III only (D) II and III only

|D |Speed depends on Kelectron = Ephoton – φ, which depends on f (Ephoton = |

| |hf) and nature of photoelectric surface (φ) |

6. If photons of light of frequency f have momentum p, photons of light of frequency 2f will have a momentum of

(A) 2p (B) ½p (C) p (D) 4p

|A |p = h/λ and λ = c/f ∴ p = hf/c (p ∝ f) |

| |If f is doubled than p is also doubled |

7. Light of a particular wavelength is incident on a metal surface, and electrons are emitted from the surface as a result. To produce more electrons per unit time but with less kinetic energy per electron, one should

(A) Increase the intensity and decrease the wavelength.

(B) Increase the intensity and the wavelength.

(C) Decrease the intensity and the wavelength.

(D) Decrease the intensity and increase the wavelength.

|B |More electrons = more intense light; reduced kinetic energy = reduced |

| |frequency or increased wavelength |

8. If the momentum of an electron doubles, its de Broglie wavelength is multiplied by a factor of

(A) ¼ (B) ½ (C) 1 (D) 2

|B |p = h/λ ∴ p ∝ 1/λ |

| |If p is doubled, then λ has to be half |

Questions 9-14 The diagram shows the energy levels for hydrogen gas.

[pic]

9. What is the energy, in eV, of a photon emitted by an electron as it moves from the n = 6 to the n = 2?

(A) 0.38 eV (B) 3.02 eV (C) 3.40 eV (D) 13.60 eV

|B |E6 = -0.38 eV and E2 = -3.40 eV |

| |ΔE = E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV |

10. The energy of the photon (in J) is closest to

(A) 6.1 x 10-20 J (B) 4.8 x 10-19 J

(C) 5.4 x 10-19 J (D) 2.2 x 10-18 J

|B |3.02 eV x 1.6 x 10-19 J/1 eV = 4.8 x 10-19 J |

11. What is the frequency of the emitted photon?

(A) 9.2 x 1013 s-1 (B) 7.3 x 1014 s-1

(C) 8.2 x 1014 s-1 (D) 3.3 x 1015 s-1

|B |E = hf |

| |f = 4.83 x 10-19 J/6.63 x 10-34 J•s = 0.7 x 1015 s-1 |

12. What is the wavelength of the emitted photon?

(A) 4.1 x 10-7 m (B) 3.3 x 10-7 m

(C) 5.4 x 10-7 m (D) 5.0 x 10-7 m

|A |c = fλ |

| |λ = c/f = 3 x 108 m/s/7.3 x 1014 s-1 = 0.4 x 10-6 m |

13. What is the minimum amount of energy needed to ionized an electron that is initially in the n = 6 energy level?

(A) 13.22 eV (B) 5.12 eV

(C) 0.38 eV (D) 0.16 eV

|C |Ionization occurs when n = ∞ |

| |ΔE = E∞ – E6 = 0 – (-0.38 J) = 0.38 J |

14. A photon having energy of 9.4 eV strikes a hydrogen atom in the ground state. Why is the photon not absorbed by the hydrogen atom?

(A) The atom's orbital electron is moving too fast

(B) The photon striking the atom is moving too fast.

(C) The photon's energy is too small.

(D) The photon is being repelled by electrostatic force.

|C |9.4 eV is not enough energy to change n |

| |(the minimum is -3.40 eV – (-13.60 eV) = 10.2 eV) |

15. Electrons with energy E have a de Broglie wavelength of approximately λ. In order to obtain electrons whose de Broglie wavelength is 2λ, what energy is required?

(A) ¼E (B) ½E (C) 2E (D) 4E

|A |E = p2/2m and p = h/λ → E = h2/2mλ2 ∴ E ∝ 1/λ2 |

| |If λ is doubled then E is ¼ |

Questions 16-18 Use the graphs to answer the questions.

(A) (B)

(C) (D)

16. Which graph shows the de Broglie wavelength of a particle versus the linear momentum?

|B |p = h/λ → p ∝ 1/λ (when p is large then λ is small and visa versa ∴ B) |

17. Which graph shows the maximum kinetic energy of the emitted electrons versus the frequency of the light?

|A |Kelectron = hfphoton – φ, which is the equation for a straight line with a|

| |negative y-intercept ∴ A |

18. Which graph shows the total photoelectric current versus the intensity of the light for a fixed frequency above the cutoff frequency?

|D |Intensity (# of photons) ∝ current (# of electrons)— intensity and current|

| |increase together |

Questions 19-20 The spectrum of visible light emitted during transitions in excited hydrogen atoms is composed of blue, green, red, and violet lines

19. What characteristic determines energy carried by a photon?

(A) amplitude (B) phase

(C) frequency (D) velocity

|C |E = hf ∴ frequency determines amount of energy |

20. Which color is associated with the greatest energy change?

(A) blue (B) red (C) green (D) violet

|D |E = hf ∴ greatest energy requires highest frequency—violet has the highest|

| |frequency |

Practice Free Response

1. The fusion of a proton (11p = 1.007276 u) and neutron

(10n = 1.008665 u) produces deuterium, 21H (2.013553 u).

a. Write a nuclear equation.

|11p + 10n → 21H |

b. Calculate the change in mass in u.

|Δm = mH-2 – mp – mn |

|2.013553 – 1.007276 + 1.008665 = -0.002388 u |

c. Calculate the change in mass in kg (1 u = 1.66 x 10-27 kg)

|-0.002388 u x 1.66 x 10-27 kg/u = -3.96 x 10-30 kg |

d. Calculate the energy in joules released.

|E = mc2 = (-3.96 x 10-30 kg)(3 x 108 m/s)2 = -3.57 x 10-13 J |

e. If the deuterium absorbs all of the energy from part d in the form of kinetic energy, what is the speed of the deuterium particle?

|m = 2.013553 u x 1.66E-27 kg/u = 3.34E-27 kg |

|K = ½mv2 → 3.57E-13 = ½(3.34E-27)v2∴v = 1.46E7 m/s |

2. The diagram shows the lowest four energy levels of an atom. An electron in n = 4 state makes a transition to n = 2.

[pic]

Determine

a. the energy of the emitted photon in eV.

|E4 = -B/n2 = -(54.4 eV)/42 = -3.4 eV |

|ΔE = E2 – E4 = -13.6 eV – (-3.4 eV) = -10.2 eV |

b. the wavelength of the emitted photon.

|E = 1240 eV•nm/λnm |

|λnm = 1240 eV•nm/10.2 eV = 122 nm |

c. the momentum of the emitted photon.

|p = h/λ |

|p = (6.63 x 10-34 J•s)/(122 x 10-9 m) = 5.44 x 10-27 kg•m/s |

The photon is incident on silver, which emits a photoelectron (m = 9.11 x 10-31 kg) with wavelength λ = 5.2 x 10-10 m. Determine

d. the momentum of the photoelectron in kg•m/s.

|p = h/λ = 6.63 x 10-34 J•s/5.2 x 10-10 m |

|p = 1.3 x 10-24 kg•m/s |

e. the kinetic energy of the photoelectron in eV.

|K = p2/2m = (1.3 x 10-24)2/2(9.11 x 10-31) = 8.9 x 10-19 J |

|8.9 x 10-19 J x 1 eV/1.6 x 10-19 J = 5.6 eV |

f. the work function of silver.

|Kelectron = Ephoton – φ |

|5.6 eV = 10.2 eV – φ ∴ φ = 4.6 eV |

Unit 6 Answers (Don't look until after you have tried the problem)

|1 |D |Strong nuclear forces > electric repulsion |

|2 |A |It takes energy to remove an electron from neutral atom ∴ separate |

| | |parts have more total mass |

|3 |C |Negative charge attracts + alpha particle |

|4 |A |Positive charge attracts – beta particle |

|5 |B |Gamma is uncharged ∴ not attracted to + or – charge |

|6 |B |Nonspontaneous ∴ energy (mass) is added to reactants form products |

|7 |C |Equal impulse (FΔt) at separation ∴ equal momentum |

|8 |B |Same momentum, but He has less mass ∴ greater velocity |

|9 |B |mHevHe = mThvTh, vHe > vTh and v is squared in K ∴ KHe > KTh |

|10 |C |31H → 0-1β + 32He |

|11 |B |21084Po → 42α + 20682Pb |

|12 |C |146C + 0-1e → 145B |

|13 |C |10n + 168O → 157N + 21H |

|14 |C |90 years = 3 t½ (16 g → 8 g → 4 g → 2 g) |

|15 |A |16 g → 8 g = 1 t½ (30 yr) |

|16 |D |1 → ½ → ¼ (2 t½) ∴ 16 yr/2 = 8 yr |

|17 |B |Shorter t½ equals more radioactive |

|18 |C |γ has no mass or charge ∴ it can penetrate matter easily |

|19 |Subatomic particle found in the nucleus |

| |Combination of protons and neutrons |

| |Atomic number = number of protons |

| |Mass number = number of protons + neutrons |

| |same Z value, but different A value |

|20 |12.0 u x 1.66 x 10-27 kg/1 u = 1.99 x 10-26 kg |

| |E = mc2 = (1.99 x 10-26 kg)(3 x 108 m/s)2 = 1.79 x 10-9 J |

|21 |235 |92 |143 |92 |235 |

|22 |Δm = 2(mp) + 2(mn) – mHe |

| |Δm = 2(1.007276) + 2(1.008665) – 4.001504 = 0.030378 u |

| |0.030378 u x 1.66 x 10-27 kg/1u = 5.04 x 10-29 kg |

| |E = Δmc2 = (5.04 x 10-29 kg)(3 x 108 m/s)2 = 4.54 x 10-12 J |

| |4.54 x 10-12 J/4 = 1.13 x 10-12 J |

|23 |42He |4 |2 |low |

| |0-1e |0 |-1 |medium |

| |γ |0 |0 |high |

|24 |10B + n → α + 7Li |1H + n → γ + 2H |50Cr + n → γ + 51Cr |

|25 |Δm = mHe + mTh - mU |

| |Δm = 4.002602 u + 234.055381 u – 238.050784 u = 0.007199 u |

| |0.007199 u x 1.66 x 10-27 kg/1 u = 1.20 x 10-29 kg |

| |Δm > 0 ∴ reaction is not spontaneous. |

| |E = mc2= (1.20 x 10-29 kg)(3 x 108 m/s)2 = 1.08 x 10-12 J |

|26 |12 days = 3 t½ → (1/8)1.776 g = 0.222 g |

|27 |4.00 u x 1.66 x 10-27 kg/1 u = 6.64 x 10-27 kg |

| |E = mc2= (6.64 x 10-27 kg)(3 x 108 m/s)2 = 5.98 x 10-10 J |

|28 |226 |86 |140 |86 |226 |

|29 |Δm = 1mp + 2mn – mH-3 |

| |Δm = (1.007276) + 2(1.008665) – (3.015500) = 0.009106 u |

| |Δm = 0.009106 u x 1.66 x 10-27 kg/1 u = 1.51 x 10-29 kg |

| |BE = mc2 = (1.51 x 10-29 kg)(3 x 108 m/s)2 = 1.36 x 10-12 J |

| |BE/A = 1.36 x 10-12 J/3 = 4.53 x 10-13 J/nucleon |

|30 |27Al + α→ 1n + 30P |9Be + α → n + 12C |

|31 |22688Ra → 22286Rn + 42He |

| |221.97036 + 4.001504 – 225.97709 = -0.005226 u |

| |-0.005226 u x 1.66 x 10-27 kg/1 u = -8.675 x 10-30 kg |

| |E = mc2= (8.678 x 10-30 kg)(3 x 108 m/s)2 = 7.81 x 10-13 J |

| |mRavRa = mαvα → (222)(4) = (4)(222) |

| |Kα/KRa = ½mαvα2/½mRavRa2 = (4)(222)2/(222)(4)2 = 222/4 = 55.5 |

| |E = ½mv2 |

| |(.98)(7.81 x 10-13 J) = ½(4 x 1.67 x 10-27 kg)v2 ∴ v = 1.51 x 107 m/s |

| |λ = h/mv = 6.63 x 10-34 J•s/(4 x 1.67 x 10-27 kg)(1.51 x 107 m/s) |

| |λ = 6.57 x 10-15 m |

|32 |The loss of 7/8 activity means that 1/8 remains It takes 3 half-lives to |

| |reach 1/8 (1 → 1/2 → 1/4 → 1/8) 3 x 14 = 42 days |

|33 |f |λ |E |m |p |I |

| |violet |red |violet |violet |violet |? |

|34 |c = fλ → 3 x 108 m/s = f(1.54 x 10-10 m) ∴ f = 1.95 x 1018 s-1 |

| |E = hf = (6.63 x 10-34 J•s)(1.95 x 1018 s-1) = 1.29 x 10-15 J |

| |E = mc2 → 1.29 x 10-15 J = m(3 x 108 m/s)2 ∴ m = 1.44 x 10-32 kg |

| |p = h/λ = (6.63 x 10-34 J•s)/(1.54 x 10-10 m) = 4.31 x 10-24 kg•m/s |

|35 |E = mc2 = 2(1.67 x 10-27 kg)(3.0 x 108 m/s)2 = 3.0 x 10-10 J |

| |E = hf → f = E/h = 3.0 x 10-10 J/6.63 x 10-34 J•s = 4.5 x 1023 s-1 |

| |c = fλ → λ = c/f = 3 x 108 m/s/4.5 x 1023 s-1 = 6.7 x 10-16 m |

| |p = h/λ = (6.63 x 10-34 J•s)/(6.7 x 10-16 m) = 1.0 x 10-18 kg•m/s |

|36 |B |E∝ – E1 = 0 – (-13.6 eV/12) = 13.6 eV |

|37 |D |The energy gap between 1 and 2 is ¾ of the total energy between |

| | |ground state and ionization |

|38 |A |Red would have the least energy ∴ the electron would come from the |

| | |energy level closest to 2 |

|39 |D |4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 → 1 |

|40 |B |Greater wavelength equals less energy (E = hc/λ) ∴ B |

|41 |A |Greater frequency equals more energy (E = hf) ∴ A |

|42 |D |red < orange < yellow < green < blue < violet ∴ blue |

|43 |C |300-nm light has more energy than 400-nm light; the extra energy |

| | |becomes electron kinetic energy |

|44 |A |Brighter (intense) has more photons, but not more energetic |

|45 |B |λ = h/mv → λ ∝ 1/v ∴ slower = longer wavelength |

|46 |B |λ = h/mv → λ ∝ 1/m ∴ lighter = longer wavelength |

|47 |C |λ = h/p → λ ∝ 1/p∴ same p = same wavelength |

|48 |c = fλ → 3 x 108 m/s = f(750 x 10-9 m) ∴ f = 4.0 x 1014 s-1 |

| |E = hf = (6.63 x 10-34 J•s)(4.0 x 1014 s-1) = 2.7 x 10-19 J |

| |E = mc2 → 2.7 x 10-19 J = m(3 x 108 m/s)2 ∴ m = 2.9 x 10-36 kg |

| |p = h/λ = (6.63 x 10-34 J•s)/(750 x 10-9 m) = 8.8 x 10-28 kg•m/s |

|49 |E2 = -B/n2 = -13.6 eV/22 = -3.4 eV |

| |E3 = -B/n2 = -13.6 eV/32 = -1.5 eV |

| |ΔE = -1.5 eV – (-3.4 eV) = 1.9 eV |

| |Ephoton = (1240 eV•nm)/λnm → 1.9 eV = 1240/λnm ∴ λnm = 653 nm |

|50 |4 → 3 |4 → 2 |4 → 1 |3 → 2 |3 → 1 |2 → 1 |

| |-12 eV |-32 eV |-140 eV |-20 eV |-128 eV |-108 eV |

|51 |20 eV, 128 eV, 108 eV |

| |E1 = -13.6 eV/12 = -13.6 eV, E2 = -13.6 eV/22 = -3.4 eV |

| |ΔE = En-high – En-low = -3.4 – (-13.6) = 10.2 eV |

| |Ephoton = 1240/λnm → λnm = 1240/10.2 = 122 nm |

| |ΔE = E∞ – En = 0 – (-B)/n2 = 13.6 eV/22 = 3.4 eV |

|52 |λnm = 1240 eV•nm /φeV = 1240 eV•nm /3.10 eV = 400 nm |

|53 |E = 1240/λnm = 1240 eV•nm/240 nm = 5.17 eV |

| |K = Ephoton - φ = 5.17 eV – 2.40 eV = 2.77 eV |

|54 |Vstop = Kelectron = 1.64 eV |

| |Ephoton = 1240/λnm = 1240 eV•nm/230 nm = 5.39 eV |

| |φ = Ephoton – K = 5.39 – 1.64 = 3.75 eV |

|55 |E = 1240 eV•nm/500 nm = 2.48 eV |

| |K = Ephoton - φ = 2.48 eV – 2.28 eV = 0.20 eV |

| |φ = 1240 eV•nm/λnm = 1240 eV•nm/2.28 nm = 544 nm |

| |0 eV, the energy of the photon is less than φ |

|56 |500 nm x 10-9 m/1 nm = 500 x 10-9 m |

| |p = h/λ = 6.63E-34/500E-9 |p = h/λ = 6.63E-34/500E-9 |

| |p = 1.3E-27kg•m/s |p = 1.3E-27 kg•m/s |

| |m = p/c = 1.3E-27/3E8= 4.4E-36 kg |9.1E-31 kg |

| |3E8 m/s |v = p/m = 1.3E-27/9.1E-31 |

| | |v = 1.4E3 m/s |

| |f = c/λ = 3E8/500E-9 = 6E14 |f = v/λ = 1.4E3/500E-9= 2.8E9 |

| |E = hf = (6.63E-34)(6E14) = 4E-19 J |E = ½mv2= ½(9.1E-31)(1.4E3)2 |

| | |E = 8.9E-25 J |

| |4E-19 J x 1 eV/1.6E-19 J = 2.5 eV |8.9E-25 J x 1/1.6E-19 J = 5.6E-6 eV |

Practice Multiple Choice (No calculator)

|1 |C |Spontaneous loses energy (mass) ∴ MX > MY + MZ |

|2 |D |21482Pb → 0-1β + xyZ: 214 = 0 + x ∴ x = 214, 82 = -1 + y ∴ y = 83 |

|3 |D |Taking apart takes energy ∴ mBE is added to the reactants |

|4 |B |t½ = 30 min and 3 half lives: 4000 → 2,000 → 1000 → 500 |

|5 |D |Speed depends on Kelectron = Ephoton – φ, which depends on f (Ephoton |

| | |= hf) and nature of photoelectric surface (φ) |

|6 |A |p = h/λ and λ = c/f ∴ p = hf/c ∝ f: 2 x f = 2 x p |

|7 |B |More electrons = more intense |

| | |reduce K = reduce f or increase λ |

|8 |B |p = h/λ ∴ p ∝ 1/λ: 2 x p = ½ λ |

|9 |B |E6 = -0.38 eV and E2 = -3.40 eV |

| | |ΔE = E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV |

|10 |B |3.02 eV x 1.6 x 10-19 J/1 eV = 4.8 x 10-19 J |

|11 |B |E = hf → f = 4.83 x 10-19 J/6.63 x 10-34 J•s = 0.7 x 1015 s-1 |

|12 |A |c = fλ → λ = c/f = 3 x 108 m/s/7.3 x 1014 s-1 = 0.4 x 10-6 m |

|13 |C |Ionization occurs when n = ∞ |

| | |ΔE = E∞ – E6 = 0 – (-0.38 J) = 0.38 J |

|14 |C |9.4 eV is not enough energy to change n |

| | |(the minimum is -3.40 eV – (-13.60 eV) = 10.2 eV) |

|15 |A |E = p2/2m and p = h/λ → E = h2/2mλ2 ∴ E ∝ 1/λ2 |

| | |If λ is doubled then E is ¼ |

|16 |B |p = h/λ → p ∝ 1/λ (inverse relation ∴ B) |

|17 |A |Kelectron = hfphoton – φ, straight line with negative y-intercept ∴ A |

|18 |D |Intensity (# of photons) ∝ current (# of electrons)— intensity and |

| | |current increase together |

|19 |C |E = hf ∴ frequency determines amount of energy |

|20 |D |E = hf ∴ greatest energy requires highest frequency—violet has the |

| | |highest frequency |

Practice Free Response

|1 |11p + 10n → 21H |

| |Δm = mH-2 – mp – mn |

| |2.013553 – 1.007276 + 1.008665 = -0.002388 u |

| |-0.002388 u x 1.66 x 10-27 kg/u = -3.96 x 10-30 kg |

| |E = mc2 = (-3.96 x 10-30 kg)(3 x 108 m/s)2 = -3.57 x 10-13 J |

| |m = 2.013553 u x 1.66E-27 kg/u = 3.34E-27 kg |

| |K = ½mv2 → 3.57E-13 = ½(3.34E-27)v2∴v = 1.46E7 m/s |

|2 |E4 = -B/n2 = -(54.4 eV)/42 = -3.4 eV |

| |ΔE = E2 – E4 = -13.6 eV – (-3.4 eV) = -10.2 eV |

| |E = 1240 eV•nm/λnm → λnm = 1240 eV•nm/10.2 eV = 122 nm |

| |p = h/λ = (6.63 x 10-34 J•s)/(122 x 10-9 m) = 5.44 x 10-27 kg•m/s |

| |p = h/λ = 6.63 x 10-34 J•s/5.2 x 10-10 m = 1.3 x 10-24 kg•m/s |

| |K = p2/2m = (1.3 x 10-24)2/2(9.11 x 10-31) = 8.9 x 10-19 J |

| |8.9 x 10-19 J x 1 eV/1.6 x 10-19 J = 5.6 eV |

| |Kelectron = Ephoton – φ → 5.6 eV = 10.2 eV – φ ∴ φ = 4.6 eV |

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