PHYSICS 30 YEAR-END REVIEW

Physics 30 Workbook - Review

Review p. 1

PHYSICS 30

YEAR-END REVIEW


Review p. 2

UNIT 1 (CONSERVATION LAWS) SUMMARY PAGE NEWTON’S LAWS CONS OF MECH. ENERGY CONS OF MOMENTUM WORK

MOMENTUM AND IMPULSE ELASTIC COLLISIONS ENERGY

If F is constant: WF = ΔE = F d cos θ Units: J where θ is angle btw F and d W > 0: F is helping motion W < 0: F is resisting motion W = 0: F and d are ⊥ If F is changing:

WF = Area under F-d graph

1st Law (constant velocity) - balanced forces ( 0=netF

!)

2nd Law: (acceleration)

amFnet!!

= 21 FFFnet!!!

+= 3rd Law: (reaction forces) If A exerts an action F on B, then B exerts an equal but opposite F on A

1. Kinetic Energy - speed Ek = 0.5 mv2 Units: J

2. Grav. Potential Energy - height Epg = m g h Units: J where h > 0 if above ref height h = 0 at ref height h < 0 below ref height 3. Elastic Potential Energy - stretch Es = 0.5 kx2 Units: J

where k is spring constant (N/m) x is length stretched or compressed from rest length

Conditions: - no friction (heat and sound) - only Fg and Fs do work - only Ek and Ep can change 1. Total energy remains constant ETi = ETf Epgi + Eki = Epgf + Ekf

2. If one energy increases, another decreases by same amount ΔEpg = −ΔKE

Power - rate at which work is done

tE

tWP Δ

== Units: Watts (W)

1-D: (Ref system)

1. Total momentum stays constant

TT pp ʹ=!!

2. Equal but opposite impulses 21 pp

!!Δ−=Δ

2-D:

• If at right angles, use tail-to-tip

• If not at right angles: 1. Find all components 2. State cons of p

TxTx pp ʹ=!!

TyTy pp ʹ=

!!

No loss in energy due to heat / sound. TT pp ʹ=

!!

EkTi = EkTf

Momentum: vmp

!!= Units: kg⋅m/s

Impulse: (constant force) vmptFnet

!!!Δ=Δ=Δ

Units: N⋅s or kg⋅m/s Cushioning:

Based on tvmFnetΔ

= ,

- increases the time to change the momentum - this decreases the force (inverse relationship)

Impulse for a changing net force: Find the area under the F-t graph.

If the line is above the t-axis, impulse is positive If the line is below the t-axis, the impulse is negative.


Review p. 3

UNIT 1: CONSERVATION LAWS

1. An object is being pushed upward with a force of 400 N and as a result, it is moving at a constant speed of 20.0 km/h. If the air resistance force on it is 170 N, then determine: a) the mass of the object b) the time it takes this object to travel 50.0 m 2. A 24.0 kg crate is being pushed forward along a level surface with a force of 380 N. As a result, it accelerates forward at 1.90 m/s2. a) Determine the friction force. b) If its initial speed is 16.0 m/s, then determine the distance travelled in 0.700 minutes. 3. A rock is thrown at an angle from a height of 1.60 m with a speed of 8.0 m/s. What is the speed of the object when it lands on the ground below? Use energetics techniques. 4. A cart is initially at a height H and moving at a speed of 6.1 m/s. When it moves to a height of 4.0 m, then it is moving at an speed of 9.5 m/s. If the system is conservative, then determine the initial height H. 5. A 6.0 kg ball, rolling East at 4.0 m/s, collides with an 8.0 kg ball rolling West at 11.0 m/s. After the collision, the 6.0 kg ball is moving West at 7.0 m.s. Assume a level surface and no friction. a) Determine the impulse on (i) the 6.0 kg ball (ii) the 8.0 kg ball b) Determine the velocity of the 8.0 kg ball after the collision. c) If the collision acted for 79.0 ms, determine the average net force on each ball. d) Is the collision elastic? Justify your answer using calculations. 6. A 180 kg object, moving South at 20.0 m/s, collides with a 130 kg object moving East at 26.0 m/s. If they stick together when they collide, then determine their velocity (magnitude and direction) after the collision. Assume no friction and a level surface. 7. A changing net force is illustrated by the graph shown. This force acts on a 72.0 kg object, which is initially at rest on a level, frictionless surface. Determine the magnitude of the object’s change in velocity. 8. A 11.0 kg gun, initially at rest, fires out a 40.0 g bullet at a velocity of 510 m/s. What is the recoil velocity of the gun?

Fnet (N)

t (s)

6.2

12 25


Review p. 4

9. A 26.0 g object has a kinetic energy of 400 mJ. What is the momentum of the object? 10. A 2.70 kg object, moving East at 6.50 m/s, collides with a stationary 4.00 kg object. After the collision, the 2.70 kg object is moving at 2.00 m/s at an angle of 51.0° S of E. What is the velocity of the 4.00 kg object (magnitude and direction) after the collision? 11. A bomb, initially at rest, explodes into three pieces (along a plane). A 12.0 kg piece is moving East at 41.0 m/s, while a 5.00 kg piece is moving at a velocity of 62.0 m/s at 73.0° S of W. Determine the velocity (magnitude and direction) of the 3rd piece after the explosion. SOLUTIONS

1. a) Balanced forces: Fg = 230 N ; m = 23.4 kg ; b) Using tdv = , t = 9.00 s

2. a) Uniform a (Unbalanced F.s): Using maFFF fnet =−=!

, Ff = 337 N backwards

b) Using 25.0 tatvd i!!!

+=Δ , d!

Δ = 2.35 km

3. Cons of E: gifi EpEkEk += (9.81) (1.60) + 0.5 (8.0)2 = 0.5 vf2 vf = 9.8 m/s

4. Cons of E: gifgfi EpEkEpEk +=+ H = 6.7 m

5. a) Impulse on 6.0 kg = vm !Δ = ( )if vvm !!− = 66 N⋅s West ; Impulse on 8.0 kg = 66 N⋅s East

b) Cons of p: TT pp ʹ=!! ⇒ (6)(4) + (8)(−11) = (6.0)(−7) + 8v ⇒ v = 2.8 m/s West

c) Equal but opposite forces (3rd Law); Ft = mΔv ⇒ F = mΔv / t = 8.4 × 102 N d) EkTi = 532 J ; EkTf = 177.25 J ; Since EkTi ≠ EkTf, it is not elastic

6. Sketch a tail-to-tip diagram: 12pʹ = 4938 kgm/s ; 12vʹ = 15.9 m/s ; θ = 43.1° E of S (46.8° S of E)

7. Impulse = Area = 0.5 (12)(6.2) + (6.2)(13) = 117.8 kgm/s ; Using Impulse = m Δv, Δv = 1.6 m/s

8. Cons of p: TT pp ʹ=!! ⇒ 0 = m1 v1ʹ + m2 v2ʹ ⇒ 2vʹ = 1.85 m/s backwards

9. Using Ek = 0.5 mv2 , v = 5.547 m/s ; Using p = mv , p = 0.144 kgm/s 10. Before: p1 = 17.55 kgm/s ; p1x = 17.55 kgm/s ; p1y = 0 After: p1ʹ = 5.40 kgm/s ; p1xʹ = 3.3983 kgm/s ; p1yʹ = −4.1966 kgm/s

Using TxTx pp ʹ=!! , p2xʹ = 14.15 kgm/s ; Using TyTy pp ʹ=

!! , p2yʹ = 4.1966 kgm/s Solving the triangle, 2pʹ = 14.76 kg⋅m/s ; v2ʹ = 3.69 m/s ; θ = 16.5° N of E (or 73.5° E of N) 11. Before explosion: pT = 0 After explosion: p1xʹ = 492 kgm/s , p1yʹ = 0 ; p2xʹ = −90.64 kgm/s , p2yʹ = −296.45 kgm/s

Using TxTx pp ʹ=!! , p3xʹ = −401.36 kgm/s ; Using TyTy pp ʹ=

!! , p3yʹ = 296.45 kgm/s Solving the triangle, 3p ʹ = 498.97 kg⋅m/s ; v3ʹ = 62.4 m/s ; θ = 36.5° N of W (or 53.5° W of N)


Review p. 5

UNIT 2 (ELECTRICITY) SUMMARY PAGE ELECTROSTATICS ELECTRIC FIELDS PARALLEL PLATES ELECTRIC FORCES

Electric Charge (+) = deficit of electrons (−) = excess of electrons

Fundamental Law Opposite charges attract Like charges repel

Induced Separation of Charge (attracting a neutral object) - if net − is brought near, e- are repelled and induces + charge - if net + is brought near, e- are attracted and induces − charge

Methods of Charging

1. Charging by Friction (insulators) - when rubbed, e- get excited - object higher on electrostatics series strips e- off other object and gains a net − charge - objects gain opposite charges

2. Charging by Conduction (Contact) - when two conductors touch, they pursue balanced (like) charges

3. Charging by Induction - when a charged object is brought near a grounded neutral object, and then the ground is removed - neutral gains the opposite charge

Coulomb’s Law (2 point charges)

221

rqqkFe =

Coulomb’s Exp’t (Torsion balance)

1. Fe vs r (inverse-square) MV: r RV: Fe (using θ) Controls: q1, q2

2. Fe vs q (direct relat.) MV: q1 RV: Fe (using θ) Controls: q2 and r

To halve the charge, touch a metal sphere with an identical neutral sphere

Uniform electric field from + to − + − Potential Difference:

qEV Δ

= Units: Volts (V)

When a charge moves freely through a V: ΔEk = +ΔE (if speeds up) ΔEk = −ΔE (if slows down) To calculate field between plates:

dVE =

! Units:

mV

where V and d are between the 2 plates

DIR: (Use positive test charge) If q is +, E

! is in same DIR as eF

!

If q is −, opposite directions MAG: 1. Small charge in the field

qFE e=

! Units:

CN

2. Central charge that makes the field

21

rqkE =

!

Field Diagrams

E!

goes from positive to negative (away from +, towards −)

Distribution of charge around conductors: For a hollow sphere, the distribution of charge is uniform. The electric field outside the sphere is uniform, but inside the sphere, it is zero. For asymmetrical hollow objects, the charge (and field) is strongest where an object comes to a point. The field is still zero inside the object.

Millikan’s Oil-Drop Exp’t

Suspended oil drop: (1st law)

ge FF =

mgEq =

All charges are multiples of e:

q = n e

Current

Current is the rate that charge flows through a device / region

tqI =

where q = ne Remember that electron flow always goes from the negative terminal to the positive terminal of a power supply. (i.e. It is attracted to the positive terminal)


Review p. 6

UNIT 2: ELECTRICITY 1. When a balloon is rubbed by fur, it gains a negative charge. Then, when it is brought near a wall, it is attracted and remains stuck to the wall. Explain these observations. 2. Two identical metal spheres, one with a charge of −72.3 mC and one with a charge of +48.1 mC, are briefly touched together and then moved to a distance of 17.0 cm apart. Determine the electric force acting on each charge. 3. The electric force between two charges is 120 N. The distance between the charges is then tripled (× 3), while both charges are halved. Determine the new electric force between them. 4. One point charge has a charge that is 4 times bigger than another charge. When they are placed 7.00 m apart, they experience an electric force of 2.90 N. Determine each charge. 5. Consider Determine the magnitude and direction of the net electric force on q1. 6. Two 5.5 g pith balls gain equal charges and are suspended from 11 cm strings, as shown. Determine: a) the electric force on each ball b) the charge of each ball (magnitude) 7. The electric field at P is 675 N/C directed towards the West. Determine the electric field and the electric force (mag and dir) if: a) an alpha particle is located at position P, and b) an electron is located at position P.

8. Consider

If the net electric field at P is zero, then find Q2. 9. Consider Determine the magnitude and the direction of the net electric field at P.

17.0 cm 10.0 cm

q1 = −185 µC q2 = 70.0 µC q3 = −160 µC

11 m 19 m

q1 = 2.6 mC

P q2 = −5.0 mC

q1 = 180 µC

67 cm 29 cm

q2 P

18° 18° 11 cm 11 cm

5.5 g 5.5 g


Review p. 7

10. Consider a Millikan-like experiment, where an 1.942 × 10-15 kg oil-drop is located between two parallel plates, 6.50 cm apart and with a potential difference of 860 V.

a) If the oil-drop is stationary, then find the charge of the oil-drop, then find the number of electrons in excess or deficit. b) If the charge of the oil-drop is +3.20 × 10-18 C, then find the acceleration of the drop. 11. Two parallel plates are 12 cm apart. A particle with 14 electrons in deficit is located 5.0 cm from the negative plate. Determine how much work is required to move the particle to the positive plate if the potential difference between the plates is 600 kV. 12. An alpha particle, moving at 2.00 × 105 m/s, is accelerated through a potential difference of 1200 V. The plates are 8.00 cm apart. Find its final speed v: a) energetically b) dynamically (using force and acceleration) Ignore the weight of the particle.

13. An electron enters a perpendicular 22.0 V/m electric field at a speed of 6.10 × 105 m/s. If it takes 1.40 µs to travel through the electric field, then determine: a) the length of the parallel plates b) the vertical deflection c) the final velocity (magnitude and direction) of the electron as it exits the field

6.50 cm 860 V 1.942 × 10−

15 kg

2.00 × 105 m/s v ?

8.00 cm

1200 V

e- 5.10 × 106 m/s

Positive plate

Negative plate


Review p. 8

SOLUTIONS

1. Charging by friction: When the balloon is rubbed by fur, the heat generated excites the electrons. Since the balloon is higher on the electrostatic series, it has a stronger hold on electrons and it pulls the electrons off the fur. Thus, it gains an excess of electrons. Induced separation of charge: When the balloon is brought near the wall, it repels the electrons in the wall (like charges). This induces a positive charge in the wall, and so, they attract (opposite charges)

2. After contact, q = 0.5 (−72.3 mC + 48.1 mC) = −12.1 mC ; 2

2

rkqFe = = 3.76 × 109 N (repels)

3. If r × 3, then F ÷ 9 ; If q1 × 0.5, then F × 0.5 ; If q2 × 0.5, then F × 0.5 ; New Fe = 3.33 N

4. ( ) ( )2

2

2

44rkq

rqqkFe == q = 6.29 × 10−

5 C

5. Fe12 = kq1q2 / r2 = 4028.4 N ; Fe13 = kq1q3 / r2 = 3650.3 N ; Fenet = Fe12 - Fe13 = 378 N right

6. a) Tgenet FFFF!!!!

++= Solving the triangle, Fe = mg tan 18° = 1.8 × 10−2 N

b) r = 2 (0.11 sin 18°) = 0.06798 m ; Using 2

2

rkqFe = , q = 9.5 × 10−

8 C

7. a) E = 675 N/C West ; Fe = qE = 2.16 × 10-16 N West (since + charge, same direction as E) b) E = 675 N/C West ; Fe = 1.08 × 10-16 N East (since − charge, opposite direction to E)

8. Balanced fields: |E2| = |E1| = 3.605 × 106 N/C ; Using |E2| = kq2 / r2 , q2 = |E2| r2 / k = 3.4 × 10−5 C

9. |E1| = 1.932 × 105 N/C ; |E2| = 1.245 × 105 N/C ; Enet = 2.3 × 105 N/C at 33° E of N (or 57° N of E)

10. a) Balanced forces: Fg = Fe ⇒ mg = q |E| ⇒ mg = qV / d ⇒ q = mgd / V = 1.44 × 10-18 C Also, n = q / e = 9.00 electrons in deficit, since q is positive. b) Fg = 1.9501 × 10-14 N ; Fe = q |E| = qV / d = 4.23385 × 10-14 N ; Newton’s 2nd Law (Ref: ↑+) a = Fnet / m = (Fe - Fg) / m = 12.0 m/s2 upward

11. q = 2.24 × 10−18 C ; ΔV = 5/12 (600 kV) = 2.5 × 105 V ; VqEW Δ=Δ= = 5.6 × 10−

13 J

12. a) ΔEk = +ΔE ⇒ Ekf − Eki = qV ⇒ 0.5 mvf2 = 0.5mvi

2 + qV ; vf = 3.94 × 105 m/s

b) ⎜E⎜ = V / d = 1.50 × 104 V/m ; 2nd Law: a = Fnet / m = Fe / m = q E / m = 7.218 × 1011 m/s2 vf

2 = vi2 + 2ad ⇒ vf = 3.94 × 105 m/s

13. a) (Horiz) Constant v (a = 0): Using td

v xx

Δ= , L = 7.14 m

b) (Vert) 2nd Law: a = Fnet / m = Fe / m = q E / m = 3.8639 × 1012 m/s2 viy = 0: Δdy = viy t + 0.5 ayt2 = 3.79 m

c) Vert: Using tvv

a iyfyy

−= , vfy = 5.41 × 106 m/s down ; vfx = vix = 5.10 × 106 m/s right

Solving the vector triangle, vf = 7.43 × 106 m/s at 46.7° below the horizontal


Review p. 9

UNIT 3 SUMMARY PAGE MAGNETISM ELECTROMAGNETISM EMAG INDUCTION

I creates a magnetic field (Oersted) HAND RULES: If q is −, use left hand If q is +, use right hand Rule #1 (B around straight I)

Thumb = Dir of I Curled fingers = Dir of B Rule #2 (Solenoids)

Curled fingers = Dir of I Thumb = Dir of B inside solenoid (points to N pole) Solenoids behave like bar magnets. To strengthen the magnetic field, insert an iron core Rule #3 (Fm)

Thumb = Dir of I Straight fingers = Dir of Bext Palm = Dir of Fm Fm on wire (with I)

⊥= LIBFm

where IB!!

⊥ Units: Teslas (T)

If IB!!|| , then Fm = 0

Fm on freely-moving charge ⊥= qvBFm

where vB!!

⊥ Units: Teslas (T)

If vB!!|| , then Fm = 0

If Fm is the only force, the particle goes in uniform circular motion

Fundamental Law

Opposite poles attract Like poles repel Magnetic Fields (B)

- use the N pole of test compass

DIR = from N to S (outside magnet) in closed loops MAG = strong if field lines are close together Earth’s Magnetic Field

- geographic N ≈ magnetic South Domain Theory

- ferromagnetic material have domains Unmagnetized: - the domains are randomly arranged - the fields cancel out Magnetized: - the domains are aligned (in the same direction as the external magnetic field) Soft ferromagnetic material (e.g. iron) - gains and loses magnetism quickly

Hard ferromagnetic material (e.g. alloys) - hard to align domains - domains stay aligned longer when external field is removed

S

N

Faraday’s law - to induce I, change the flux (the number of B lines in loop)

Lenz’s Law - induced I creates a B which fights (resists) the change in flux For solenoids, the induced Fm always goes opposite to the motion of the magnet. Determine if this is attraction or repulsion, and based on this, determine the nature of the pole for the solenoid. Use LHR #2 to determine the direction of the current. For moving wires, Fm always goes opposite to the motion of the perpendicular wire moving through the field. Then, use LHR #3 to determine the direction of the e- flow through the wire.


Review p. 10

UNIT 3: ELECTROMAGNETISM 1. Explain how the South end of a magnet can stick to a fridge door.

2. a) The parallel wires shown have b) Which way will the compass deflect? opposite currents. Will they attract or repel? c) What is the direction of the magnetic d) What is the direction of the charge’s velocity? force on the proton? 3. A 7.50 cm length of wire is suspended in a 4.90 T perpendicular magnetic field. The mass of the wire is 13.0 g. a) Determine the direction of the electron flow through the wire.

b) Determine how much charge flows through the wire in 4.00 minutes.

4. When an alpha particle enters a perpendicular magnetic field at a speed of 4.92 × 105 m/s, it experiences an acceleration of 6.18 × 109 m/s2. What is the magnitude of the magnetic field strength? 5. Determine the direction of the induced electron flow through the galvanometer. a) b)

Compass

• ×

e- e-

N S p+

v

Fm q-

× × × ×

× × × ×

× × × ×

N S

G

N

S

G

S N

X

Y


Review p. 11

6. Determine the direction of: a) the motion of the wire b) the electron flow around the loop SOLUTIONS

1. When the S end of the magnet is brought near, it causes the domains to align towards the S end (in the same direction as the external magnetic field). This induces a N pole in the fridge door (closest to the magnet). Opposite poles attract.

2. a) (LHR #1) Left wire: B is clockwise; Right wire: B is counterclockwise ; Since there are like fields (same direction) in between the wires, the wires will repel b) (LHR #2) e- flows from − to + ; N pole on left, S on right ; Compass points left c) (RHR #3) Fm acts out of the page (•) d) (LHR #3) v is towards the bottom of the page (↓)

3. a) (LHR #3) Thumb = Dir of e- (from X to Y) ; Straight fingers = Bext (from N to S = left) Palm = Dir of Fm (opposite to Fg = up) b) Balanced forces: Fm = Fg LIB⊥ = mg (0.075) I (4.90) = (0.013) (9.81)

I = 0.347 A ; t = 240 s ; Using tqI = , q = 83.2 C

4. 2nd law: Fnet = ma ⇒ Fm = m a q v B⊥ = m a

( )( )( )( )smC

smkg

qvmaB

519

2927

1092.41020.3

1018.61065.6

××

××==

−

−

⊥ = 2.61 × 10−4 T

5. a) Lenz's Law: Fm on magnet is opposite to the motion (left) ; Induces a N pole on right of solenoid ; Using LHR #2, e- flows left through galvanometer b) Lenz's Law: Fm on magnet is opposite to the motion (up) ; Induces a N pole on top of solenoid ; Using LHR #2, e- flows down through galvanometer 6. a) Using LHR #3, Fm acts to the left on the wire; Lenz's Law: Wire is moving to the right b) Analyze the top wire, since it is perpendicular to the field and it sweeps through the field. Lenz's law: Fm acts upward on the wire ; LHR #3: e- flows to the left ; I is counterclockwise

× × ×

× × ×

× × ×

A coil pulled out of a magnetic field.

B • •

• •

• •

• •

•

•

•

•

•

e-


Review p. 12

UNIT 4 SUMMARY PAGE

Light Review

- a wave is a transmission of energy - the source of all waves is a vibration (provides the energy, determines the frequency) - f never changes once it leaves the source - if the medium stays the same, the speed is constant λfv = - f = # of cycles / s (Units: Hz) - period (T) is the time per cycle

f

T 1=

EMR and Maxwell’s Laws

- source = accelerating charge - creates ΔB, ΔE, ΔB, ..., moving at the speed of light - vandEB

!!!,, are always ⊥

- in order of increasing f: Power, Radio, Microwaves, IR, Visible (ROYGBIV), UV, X-rays, Gamma Speed of Light

- Galileo used lanterns - Roemer and Huygens used the eclipse of Jupiter's moon - for Michelson, he shone light on an 8-sided mirror, reflected off a distant mirror, and returned; he increased the f of rotation until he could see the light - the time it takes for light to travel to the distant mirror and back is the time for 1/8 of a revolution i.e.

TDv812

≈

Refraction

Fast to slow: - partial reflection and refraction - ray bends towards the normal Slow to fast: - partial reflection and refraction - ray bends away from the normal Snell's Law:

2

1

2

1

1

2

2

1

sinsin

vv

nn

===λλ

θθ

- frequency never changes Total Internal Reflection

Critical angle (θc): - slow to fast; θR = 90° If θi > θc, there is no refracted ray Diffraction

The longer the wavelength and the narrower the opening, the more diffraction (i.e. bending). Young's Double Slit

nLdx

nd

==θ

λsin

d = distance between slits x = distance between maxima L = distance to screen Dispersion

- for prisms, shorter wavelengths bend more; for diffraction grating, longer wavelengths bend more Polarization

- when a wave travels through a vertical slit / polarizer, the wave is entirely vertical (horiz is absorbed) - if it then travels through a horizontal polarizer, no wave gets through

Law of Reflection

- angle of incidence is equal to the angle of reflection (symmetrical)

ri θθ = - angles are always measured with respect to the normal - for plane mirrors, the image is the same size, upright, virtual, ands the same distance from the mirror Mirrors / Lenses

To find the image:

1. Draw two of the 3 rays: Parallel - focus, centre - along the same path, focus - parallel 2. Image is located where the reflected (or refracted) rays converge Real image are always inverted Virtual images are always upright Sign Convention:

h, mag + : upright h, mag − : inverted

d, f + : real d, f, − : virtual

mag > 1: image is larger mag = 1: image is same size mag < 1: image is smaller

fdd io

111=+ mag

dd

hh i

o

i =−=0

fR 2= Index of Refraction

vcn =

When n is high, the speed is slower (optically dense). When n is low, the speed is faster.

For air, n = 1


Review p. 13

UNIT 4: EMR

1. For the EMR shown, determine the period of the source vibration. 2. The carrier wave for an FM radio station is 100.3 MHz. Determine: a) the wavelength of the carrier wave b) the time it takes this wave to travel 220 km 3. Two mirrors form a 119° angle when connected. If the incident ray is 40° from the surface of mirror I, then determine the angle of reflection off of mirror II. 4. In a Michelson experiment, the 8-sided mirror must rotate at a minimum rate of 90,000 rpm to see a continuous beam of light. If the measured speed of light was 2.75 × 108 m/s, then determine the distance between the two mirrors. 5. A 60.0 cm high object is placed 80.0 cm from a concave (converging) mirror. If the real image produced is 35.0 cm high, then determine the focal length of the mirror. 6. When a 17 cm high object is placed 22 cm away from a mirror, the image produced is exactly the same size. Determine the type of mirror and its focal length. 7. A convex (diverging) mirror has a radius of curvature of 90 cm. If the distance to the object is 50 cm, then determine its magnification. 8. EMR enters a liquid (n = 1.72) at an angle of 38.0° w.r.t. the surface. When the EMR is inside the liquid, it has a wavelength of 460 nm. Determine: a) the angle of refraction b) the frequency of the EMR inside the liquid 9. The critical angle for an air - liquid interface is 44.0°. Determine the speed of the EMR in the liquid. 10. A 11.0 cm high object is placed in front of a convex (converging) lens that has a focal length of 30.0 cm. The virtual image created is located 50.0 cm from the lens. Determine the height of the image. 11. 860 nm EMR is sent through diffraction grating that is rated at 200 slits per mm. Determine the angle to the second order maximum. 12. EMR with a frequency of 6.30 × 1014 Hz is sent through diffraction grating that has 7.20 × 105 slits per metre. If the distance between the maxima is 18.1 cm, then determine the distance from the diffraction grating to the screen.

57 nm

40º

119º

I

II

38.0°


Review p. 14

13. Two microwave transmitters are placed side-by-side, each emitted EMR in phase with a frequency of 5.0 GHz. If a receiver is placed 15 cm from one transmitter and 27 cm from the other transmitter, then would there be a strong or a weak signal? Justify. SOLUTIONS

1. From the graph, λ = 38 cm ; Using λfv = , f = 1.5789 × 1016 Hz ; f

T 1= = 6.3 × 10−

17 s

2. a) v = f λ ⇒ λ = v / f = 3.00 × 108 m/s / 100.3 × 106 Hz = 2.99 m b) v = d / t ⇒ t = d / v = 220 × 103 m / 3.00 × 108 m/s = 7.33 × 10-4 s

3. Using law of reflection (θi = θr), and angles in a triangle add up to 180°, θr = 69°

4. f = 90,000 / 60 s = 1500 Hz ; T = 1 / f = 6.667 × 10−4 s ; Using

TDv812

≈ , D = 11.5 km

5. ho = 60 cm (upright) ; do = 80 cm (real) ; hi = −35 cm (inverted, since real) ;

Using o

i

o

i

dd

hh

−= , di = 46.667 cm (real) ; Using fdd io

111=+ , f = 29.5 cm (real)

6. Converging mirror ; Object is placed on centre, so R = 22 cm and f = 11 cm (real)

7. f = −45 cm (virtual focus) , do = 50 cm (real)

Using fdd io

111=+ , di = −23.68 cm (virtual) ;

o

i

dd

mag −= = 0.47 (or 47%)

8. a) θi = 52° ; Using 2211 sinsin θθ nn = , θR = 27.3°

b) Using vcn = , v2 = 1.744 × 108 m/s ; Using λfv = , f = 3.79 × 1014 Hz

9. Must go liquid to air; θR = 90° ; Using 2211 sinsin θθ nn = , n1 = 1.43956

Using vcn = , v1 = 2.08 × 108 m/s

10. ho = 11 cm (upright) ; f = 30 cm (real) ; di = −50 cm (virtual) ;

Using fdd io

111=+ , do = 18.75 cm (real) ; Using

o

i

o

i

dd

hh

−= , hi = 29.3 cm (upright)

11. d = 1 × 10−3 m / 200 = 5.0 × 10−

6 m ; n = 2 ; Using n

d θλ

sin= , θ = 20.1°

12. λ = 4.762 × 10−7 m ; d = 1.389 × 10−

6 m ; x = 0.181 m ; Using nLdx

=λ , L = 52.8 cm

13. λ = 6.0 cm ; The first is 2.5 wavelengths to the receiver, while the second is 4.5 wavelengths to the receiver. So, if a crest is sent from both, they both receive a trough. Constructive interference = strong signal.


Review p. 15

UNIT 5 SUMMARY PAGE MODELS OF THE ATOM MATH OF KEY TOPICS Billiard Ball Model (Dalton) - atom is smallest piece of matter

Cathode Rays (electrons inside atom) - deflected towards the positive plate - deflected in B using LHR #3 (EMR does not deflect)

Raisin Bun Model (Thomson) - electrons are uniformly distributed inside a positive fluid matrix

Gold Foil Experiment (Rutherford) - fired α+2 through thin gold foil - raisin bun model predicted that there would be no deflection - Results: 99.99% went through undeflected, but a few (1 in 10,000) deflected significantly

Planetary Model (Rutherford) - electrons orbit a tiny, massive, positive nucleus ; mostly empty space

Problem with Planetary Model - orbiting e- is accelerating, and thus, it should be constantly emitting energy (Maxwell’s laws) = collapse in 10−

8 s

Quantum Theory

UV Catastrophe (Blackbody radiation) - classical theory predicted that short λ EMR would be infinitely bright, but in fact, they were very dim

Quantum Hypothesis (Planck) - explained the UV catastrophe - vibrating atoms have quantized energy levels and energy is emitted as a bundle (photon) only when the energy level changes

3 types of Spectra: - continuous spectrum: hot dense objects emit all frequencies - emission spectrum: low pressure, excited gas emits only certain f - absorption spectrum: unexcited, low pressure gas absorbs certain f of light from hot dense objects

Bohr H Model (quantum and classical) - orbital radius and total energy of e- is quantized - EMR is emitted or absorbed only when e- changes orbits - explained the H emission spectrum

Bohr H Model

Emitting / Absorbing photons:

Cons of energy:

photone EE =Δ − Ionization Energy: Minimum energy to remove e- Energy to n = ∞ Explaining H spectrum: Lyman: (UV) nL = 1 Balmer: (visible) nL = 2 nu = 3 (red), 4 (Green), 5 (blue), 6 (violet) IR: nL = 3, 4, 5 ... Compton Effect (photon momentum)

- when x-rays were fired into thin C, x-rays were deflected at a lower f (lower E, longer λ) and an e- was emitted (photoelectric effect) - this showed that photons collide like particles, which means they have p

pcE = - to find the new wavelength,

( )θλ cos1−=Δmch

θ is the angle of deflection of x-ray (w.r.t. its original direction) Matter Waves (de Broglie)

- when e- were deflected off crystals, they diffracted and interfered like x-rays (behaved like waves)

mvh

ph==λ

Double-Slit Experiment for e-

When e- were fired into a double slit, they formed an interference pattern like waves. Verified wave-particle duality.

Electron Properties (Thomson / Millikan)

Undeflected = constant v: (1st law)

Fm = Fe Deflected (circular motion)

Fm = mac Suspended Oil Droplet:

Fg = Fe

Photoelectric Effect

- Classical theory predicted that all EMR would emit e-; the brighter the EMR, the faster the emission

Actual results: (explained by Einstein) - if f < f0, no photocurrent (photon energy lower than the work function) - if f = f0, minimum energy to emit e-

W = h f0 - if f > f0, photon energy is greater than W The photon is fully absorbed, and the extra energy is kept by the e- as Ek

Cons of energy

ETi = ETf

Ephoton = W + Ekmax - the brighter the EMR, the greater the photocurrent (since there are more photons being absorbed by e-) Verified by Millikan - measured KEmax using Vstop

Ekmax = q Vstop - using Ekmax vs f graph, showed that slope = h, x-int = f0, and y-int = W


Review p. 16

UNIT 5: ATOMIC THEORY

1. An electron first travels undeflected through perpendicular magnetic and electric fields. Then, it enters into a perpendicular magnetic field, where it goes into uniform circular motion. Undeflected Deflected in a Magnetic Field V = 1700 V r = 83.0 mm |E| = 2.90 × 104 N/C v = 4.50 × 106 m/s B1 ? B2 ? a) For the undeflected region, determine: (i) the distance between the parallel plates (in cm) (ii) the magnitude and direction of B1 (in mT).

b) For the region when it is deflected in a magnetic field, determine: (i) the magnitude of B2 (iii) the direction of the electron’s path (clockwise or counterclockwise). 2. A 6.0 W light bulb emits monochromatic blue light (λ = 420 nm). a) What is the frequency of the light? b) What is the total energy of 40.0 photons emitted by the light bulb? Answer in eV. c) How many photons would be emitted in 2.0 seconds, assuming 10% efficiency? 3. The minimum frequency to emit electrons from a metal is 5.50 × 1014 Hz. Determine the nature of the photocurrent (zero, low, or high) if the following incident lights are shone on the metal: a) dim vs bright red b) dim vs bright violet 4. The minimum energy to dig electrons out of a metal surface is 8.10 eV. a) What is the maximum wavelength (in nm) required to emit electrons from the metal surface? b) What is minimum potential difference to prevent photocurrent if the frequency of the incident EMR is 5.00 × 1015 Hz? 5. A fictional metal requires a back voltage of 14.00 V to stop the photocurrent when incident light of wavelength 51.0 nm is shone on it. a) What is the work function of the metal (in eV)? b) What is the maximum speed of a photoelectron if λ = 30.0 nm? c) How would Vstop change if the incident light is: (i) brighter? (ii) longer λ? 6. How fast would an alpha particle have to be moving to have the same momentum as a 75.0 MeV photon?

+ + +

− − −

e- 4.50 × 106 m/s

• • • •

• • • • B2


Review p. 17

7. In a Compton experiment, a 3.40 pm x-ray collides with a stationary electron. The deflected x-ray has a momentum of 1.33 × 10−

22 kg⋅m/s. Determine the angle of x-ray deflection. 8. In a Compton experiment, a 5.00 × 1018 Hz x-ray photon collides with a stationary electron and loses 12.0% of its energy to the electron. a) What is the speed of the electron that is emitted? b) What is the wavelength of the emitted x-ray? 9. For the Bohr hydrogen model, determine the orbital radius of an electron orbiting with a speed of 546 km/s. 10. Consider the energy diagram for a fictional atom.

a) What is the ionization energy of the atom (in Joules) if it is originally at n = 3? b) What wavelength of light is emitted when the atom drops from n = 4 to n = 2? c) Describe what happens when an unexcited atom receives: (i) a 10.0 eV photon (ii) a 14.0 eV photon (iii) a 25.0 eV photon d) How would your answer to #9c differ if it was an incident electron instead? 11. A particle with a charge-to-mass ratio of 4.88 × 109 C/kg is accelerated through a potential difference of 70.0 kV. Determine its final speed. 12. Determine the kinetic energy of an alpha particle that has a wavelength of 7.80 fm. Answer in MeV. SOLUTIONS 1. a) (i) |E| = V / d ⇒ d = V / |E| = 5.86 cm (ii) (balanced forces) Fm = Fe qvB1 = q |E| B1 = |E| / v = 6.44 mT (LHR #3) Thumb = Dir of velocity (right); Straight fingers = Dir of Bext (into the page) Palm = Dir of Fm (opposite to Fe, so downward)

b) (i) Uniform circular motion: Fnet = ma Fm = mac rmvqvB

2

=⊥ ⊥B = 3.09 × 10-4 T

(ii) Using LHR #3, Fm acts upward and so it moves counterclockwise

n = 1

n = 2

n = 3

n = 4

n = ∞

0 eV

11.5 eV

17.1 eV

19.2 eV

21.0 eV


Review p. 18

2. a) f = c / λ = 3.00 × 108 m/s / 420 × 10-9 m = 7.14 × 1014 Hz

b) Ephoton = hc / λ = (4.14 × 10-15 eVs) (3.00 × 108 m/s) / (420 × 10-9 m) = 2.957 eV ET = n Ephoton = 40 (2.957 eV) = 118 eV c) ET = P t = 12.0 J ; Ephoton = hc / λ = 4.74 × 10-19 J ; n = ET / Ephoton = 2.5 × 1019 photons

3. λ0 = 545 nm ; a) Red (700 nm) has too small of a frequency to emit electrons. No I for each. b) Violet (400 nm) has a high frequency. Dim violet creates a small I; Bright violet has a high I

4. a) W = hf0 ⇒ f0 = W / h = 1.96 × 1015 Hz ; λmax = c / f0 = 1.53 × 10-7 m = 153 nm b) Cons of E: Ephoton = W + Ekmax ; Ekmax = hf - W = 2.1016 × 10-18 J ; Vstop = Ekmax / q = 12.6 V

5. a) Ekmax = qVstop = 2.24 × 10-18 J ; f = c / λ = 5.882 × 1015 Hz Cons of E: W = hf - Ekmax = 1.66 × 10-18 J ; JW 181066.1 −×= = 10.4 eV b) f = c / λ = 1.00 × 1016 Hz ; Cons of E: Ekmax = hf - W = 4.97 × 10-18 J Using Ek = 0.5 mv2, v = 3.30 × 106 m/s c) (i) No change (intensity has no effect on Ekmax or Vstop) (ii) Longer λ = less energy per photon = less Ekmax = less Vstop required

6. eVEphoton6100.75 ×= = 1.20 × 10-11 J ; Using E = p c , p = 4.0 × 10−

20 kgm/s Same momentum ; Using p = mv, v = 6.02 × 106 m/s

7. Using λhp = , λf = 4.985 pm ; Using ( )θλ cos1−=Δ

mch , θ = 69.7°

8. a) Ex-ray = hf = 3.315 × 10-15 J ; Eke- = 0.12 (3.315 × 10-15 J) = 3.978 × 10-16 J Using Ek = 0.5 mv2 , v = 2.96 × 107 m/s b) Ex-rayʹ = 0.88 Ex-ray = 2.917 × 10-15 J ; Using E = hc / λ , λ = 6.82 × 10-11 m

9. Circ Motion: Fnet = ma Fe = mac kq2 / r2 = mv2 / r r = 8.474 × 10-10 m

10. a) Minimum energy to n = ∞ ; ΔE = E∞ - E3 = 3.9 eV = 6.24 × 10-19 J b) Cons of E: ΔEe- = Ephoton ⇒ E4 - E2 = hc / λ λ = 1.61 × 10-7 m

c) (i) Nothing. The photon does not have enough energy. (ii) Nothing. The photon does not have the exact energy (assuming it is fully absorbed) (iii) Ionization. The photon is fully absorbed, and the e- leaves with 4.0 eV of KE.

d) (i) Same as photon. The electron does not have enough energy. (ii) The atom rises to n = 2, and the incident e- leaves with 2.5 eV. (iii) The atom could absorb any energy from the incident electron. The most likely two options are ionization or up to n = 2.

11. Cons of E: EEk Δ+=Δ leads to qVmv =25.0 Vv

mq 25.0= v = 2.61 × 107 m/s

12. Using mvh

ph==λ , v = 1.278 × 107 m/s ; 2

21 mvEk = = 5.432 × 10−

13 J = 3.40 MeV


Review p. 19

UNIT 6 SUMMARY PAGE

Parts of an atom

Electron: q = −e, m = 0 Proton: q = +e, m = 1u Neutron: q = 0, m = 1 u where 1 u = 1.66 × 10−

27 kg Isotopes have the same # or p+ (i.e. same element) but different # of n Strong nuclear force holds the nucleus together. Attraction force between p-p, p-n, and n-n; strong only at very short distances. Mass-Energy Equivalence

Nuclear Mass Defect:

The total mass of a nucleus is less than the sum of the nucleon masses.

Δm = m(p+n) − mnucleus Einstein stated that the lost mass had converted to another form of energy. 2cmE Δ= Note: Energy converted from the nuclear mass defect is binding energy Radiation

Three types: 1. Alpha particles (He-4) - lowest penetration, dangerous if swallowed 2. Beta particles (e-) - medium penetration, danger if swallowed 3. Gamma particles - high f EMR, highest penetration, blocked by lead Separate these using perpendicular magnetic or electric fields.

Fission Reactions:

- a large nucleus is bombarded by a slow moving nucleus, and as a result, it splits into two smaller (radioactive) daughter nuclei and some neutrons - the neutrons released cause other fission reactions (chain reaction) Fusion Reaction:

- when small nuclei merge (at high speeds) to form a larger nucleus Particle Tracks

For cloud / bubble chambers, charged particles leave a path. You can predict the q / m / v based on the nature of deflection in a magnetic field. Neutral particles leave no path. Particle Accelerators

To penetrate the nucleus, very high particle energies are needed. Sources include: 1. Cosmic rays 2. Van de Graaf generators 3. Cyclotrons 4. Synchrotrons Particle Classification

1. Leptons - do not experience strong nuclear force; zero size and no internal structure; include e-, e+, and neutrinos

2. Hadrons - experience strong nuclear force; composed of smaller particles (quarks); include p+ and n

3. Field particles - photons Quarks and Beta Decay

Proton: uud ; Neutron: udd

Beta-negative decay: n → p+ + e- + anti-neutrino down quark converts to up quark

Beta-positive decay: p+ → n + e+ + neutrino up quark converts to down quark

Decay Reactions

Top # = Conservation of mass Bottom # = Cons of charge In alpha and beta decay, mass defect occurs. The energy is released as Ek of products and gamma radiation. When the # of p+ change, the element changes (transmutation) 1. Alpha Decay

- nucleus has too many p+ - alpha particle (He-4) is emitted 2. a) Beta-Negative Decay

- nucleus has too many neutrons - n converts to a p and an electron is emitted - for conservation of energy and momentum, an antineutrino is also emitted 2. b) Beta-Positive Decay

- p converts to a n and a positron (anti-electron) is emitted - for conservation of energy and momentum, a neutrino is also emitted 3. Gamma Decay - during most alpha and beta decays, some of the energy causes the daughter nucleus to become excited and unstable; when it comes back to ground state, it emits a gamma photon Radioactivity

- rate of emissions is affected by: 1. mass of sample (direct relat.) 2. Half-life (T1/2) - the time it takes for half of the sample to decay

( )nNN 5.00= N0 = original sample N = final sample

n is the # of half-lives:

21Ttn =


Review p. 20

UNIT 6: NUCLEAR THEORY 1. For the isotope Antimony-121, determine: a) the number of neutrons, protons, and nucleons b) the nuclear charge (in C) 2. The mass of a fictional atom is 5.80 TeV/c2. Determine this mass in atomic mass units (u). 3. Determine the binding energy for the isotope Molybdenum-98 (97.905405 u). Answer in MeV. 4. Write a balanced nuclear equation for: a) the alpha decay of Th-232 b) the beta-negative decay of Ac-228 c) the beta-positive decay of C-10 5. A fictional parent nucleus (at rest) undergoes beta decay and the mass defect experienced is 0.170 u. If the kinetic energy of the emitted beta particle is 60.0 MeV and the neutrino is 18.0 MeV, then determine the frequency of the gamma photon emitted. Ignore the energy of the daughter nucleus. 6. A fictional parent nucleus (at rest) undergoes alpha decay. The alpha particle is emitted at a speed of 8.20 × 107 m/s. If the daughter nucleus has a mass of 110 u, then determine its recoil speed. 7. A singly ionized particle travels undeflected through a velocity selector at 1.9 × 104 m/s. It then enters a uniform, perpendicular 1.0 × 10-3 T magnetic field (B2) and gets deflected into uniform circular motion. a) If the magnetic field in the velocity selector (B1) is 2.6 × 10−

4 T, then find the electric field in the velocity selector. b) If the radius of curvature in the magnetic field B2 is 0.40 m, find the mass of the particle. 8. What is the charge-to-mass ratio for a particle accelerated to a velocity of 6.0 × 106 m/s, that moves in a circular path of radius 1.8 cm perpendicular to a uniform magnetic field of 3.2 × 10−

2 T? 9. Rn-222 has a half-life of 3.82 days. a) If a sample has activity 7.10 kBq, what is its activity after 58.0 hours? b) After a 42.0 days, only 15.0 grams of a sample of Rn-222 is left. What is the original amount (in kg)? 10. The original sample of a radioisotope has a mass of 63.00 g. After 116 years, it has a mass of 3.9375 g. What is its half-life? 11. When Pu-239 merges with a neutron, it goes through nuclear fission and this results in Ba-143, another daughter nucleus, and two neutrons being emitted. What is the unknown daughter nucleus? 12. In a fusion reaction, deuterium (H-2) fuses with another nucleus and as a result, an alpha particle and a proton are created. Determine the identify of the other nucleus. 13. Sodium-22 decays to Neon-22. Describe what happens to the quarks in the nuclei, and identify the emitted particles.


Review p. 21

14. You have a radioisotope and you wish to determine the nature of the particles being emitted. Explain how you could separate and identify the different particles, using: a) a cloud chamber and a magnetic field b) materials (acting as a filter) 15. In a cyclotron, a proton enters the gap with a speed of 7.1 × 106 m/s. If it speeds up through a potential difference of 80 kV, then determine its final speed. SOLUTIONS

1. a) 51 p, 70 n, 121 nucleons b) q = 51 e = 8.16 × 10−18 C

2. E = 9.28 × 10−10 J ; Using E = mc2, m = 1.031 × 10−

26 kg = 6.21 u

3. mp+n = 1.6366 × 10−25 kg ; mnucleus = 1.62523 × 10−

25 kg ; Δm = 1.137 × 10−27 kg

2cmE Δ= = 1.0233 × 10−10 J = 6.40 × 108 eV = 640 MeV

4. a) 90232

88228

24Th Ra→ + α b) ( )89

22890228

10

00Ac Th→ + +− β υ c) ( )6

10510

10

00C B→ + +β υ

5. Δm = 2.822 ×10−28 kg ; 2cmE Δ= = 2.54 × 10−

11 J = 158.74 MeV Cons of E: γυβ EEkEkE m ++=Δ Eγ = 80.74 MeV Using E = hf, f = 1.95 × 1022 Hz

6. md = 1.826 × 10−25 kg Cons of p: TT pp ʹ=

!! dd vmvm !!+= αα0 dv

! = 2.99 × 106 m/s

7. a) (Balanced forces) Fe = Fm ⇒ q |E| = qvB1 |E| = vB1 = 4.94 N/C b) Circ Motion: Fnet = ma Fm = mac qvB2 = mv2 / r m = 3.37 × 10-27 kg

8. Circ Motion: Fnet = ma Fm = mac qvB2 = mv2 / r rBv

mq

⊥

= = 1.0 × 1010 C /kg

9. a) T1/2 = 3.82 days = 91.68 h ; n = t / T1/2 = 58 h / 91.68 h = 0.63264 half-lives N = N0 (0.5) n ⇒ N = (7.10 kBq) (0.5) 0.63264 = 4.58 kBq b) n = t / T1/2 = 10.995 half-lives ; Using N = N0 (0.5) n , N0 = 3.06 × 104 g = 30.6 kg

10. Using N = N0 (0.5) n , n = 4 half-lives ; Using n = t / T1/2 , T1/2 = 29.0 y

11. 94239

01

56143

3895

012Pu n Ba Sr n+ → + + . The unknown nucleus is Sr - 95.

12. 23

12

24

11He H He H+ → + . So, the nucleus is He-3.

13. Beta-positive decay: p+ (uud) changes to a neutron (udd). So, an up quark changes to a down quark, and in the process, a positron and a neutrino are emitted.

14. a) Electrons (beta) would obey LHR #3 and have a small radius ; α+2 obey RHR #3 and have a large radius ; Gamma rays do not deflect and leave no tracks in a cloud chamber b) Place a few sheets of paper as a barrier: α+2 do not penetrate, but e- and γ will Place a thin piece of foil / lead as a barrier and e- would not penetrate, but γ will

15. Cons of E: EEk Δ+=Δ ; qVEkEk if =− ; qVmvmv if += 22 5.05.0 ; vf = 8.1 × 106 m/s


Review p. 22

GRAPHICAL ANALYSIS 1. In Young’s double-slit experiment, a Physics 30 student shone monochromatic EMR through diffraction grating that is rated at 520 slits / mm. The distance from the diffraction grating to the screen was varied and the distance between the maxima was measured. The results are:

L (cm) 10 14 18 22 26 30 x (cm) 3.5 4.5 6.0 7.5 9.0 10.0

a) Graph the data. Determine the slope of the best-fit line. b) Using the significance of the slope, determine the wavelength of the EMR used. 2. In a photoelectric experiment, the manipulated variable is the frequency of the incident light, and the responding variable is the maximum kinetic energy of the emitted photoelectrons. The results are:

f (× 1015 Hz) 2.0 4.0 6.0 8.0 10.0 12.0 Ekmax (eV) 3.0 9.5 19.5 25.0 36.0 42.5

Graph the data. Using the line of best-fit, determine: a) the maximum wavelength of EMR to emit photoelectrons from the metal surface b) the measured value for Planck’s constant h. c) Vstop when incident light has a f = 7.0 × 1015 Hz. 3. In a Millikan experiment, an oil-droplet is suspended between parallel plates that are 11.0 cm apart. The charge of the oil droplet is varied (using x-rays) and the potential difference is adjusted until the oil droplet is suspended. The results are shown below:

OBSERVATIONS ANALYSIS

Oil droplet’s charge, q

(× 10−19 C)

Potential Difference, V

(× 102 V) q1

( )

Potential Difference, V

(× 102 V)

1.6 18.0 10 3.2 9.0 18 4.8 6.0 24 6.4 4.5 37 8.0 3.5 45

a) Explain why the analysis leads to a straight line through the origin. b) Graph the analyzed data. Determine the slope and include its units. c) Using the significance of the slope, determine the mass of the oil droplet.


Review p. 23

4. In an optics experiment, a candle is placed in front of a converging (convex) lens. The distance from the lens to the candle flame is varied and the distance from the lens to the image is measured. The results are shown below:

OBSERVATIONS ANALYSIS

Distance to object, do

(cm)

Distance to image, di

(cm) od1

( ) id1

( )

30 60 40 41 50 35 60 31 70 27

a) Graph the analyzed data. Determine the slope and the y-intercept. b) Using the line of best-fit, determine the focal length of the lens. SOLUTIONS 1. a) Slope = 0.339 (no units)

b) Graph equation: ( )Lslopex = Known equation: nLdx

=λ or dLn

xλ

=

dn

slopeλ

= ; d = 1.923 × 10−6 m ; n = 1 ; λ = 6.5 × 10−

7 m

2. a) f0 = x-intercept = 1.404 × 1015 Hz ; Using λfv = , λ = 2.1 × 10−

7 m b) h = slope = 4.0 × 10−

15 eV⋅s c) When f = 7.0 × 1015 Hz, Ekmax = 22.58 eV = 3.61 × 10−

18 J ; Vstop = 23 V

3. a) Balanced forces: ge FF = mgEq = mgdqV

= qmgdV =

qV 1∝

So, V has a direct relationship with 1/q b) Slope = 2.89 × 10−

16 C⋅V

c) Graph equation: q

slopeV = So, slope = mgd m = 2.7 × 10−

16 kg

4. a) Slope = −1.0 (no units) ; y-int = 5.0 m−

1 or 5.0 1/m

b) Graph equation: ( ) ( )int11−+= y

dslope

d oi

Known equation: fdd oi

111+−=

y-int = f1 f = 0.20 m = 20 cm

PHYSICS 30 YEAR-END REVIEW

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