2011 Quantum I Solutions

8/13/2019 2011 Quantum I Solutions

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Quantum Physics I - Solutions

_Photons, Power, Intensity, Photocurrent_

1 Number of photons emitted in 1 second

=

=

=

=

3

34 8 9

16

total energy of photons emitted in 1 second

energy of each photon

4 10 (1)

(6.63 10 )(3 10 ) / 632 10 )

1.27 10

Pt

hc

[Note: number of photons has no unit so the answer has no unit, but if the question were to

ask for number of photonsper second, then the unit is s1

.]

2 [Note: 10 W lamp means 10 W of electrical power is used up by the lamp, but not allelectrical power is converted to light power, the percentage converted to light power is givenby the efficiency of the lamp.]

(a) Given efficiency is 10%, light power produced by lamp = 10% of 10 W = 1.0 W

Number of photons produced each second by the lamp

=

=

=

=

34 8 9

18

total energy of photons emitted in 1 second


1.0(1)

(6.63 10 )(3 10 ) / (590 10 )

3.0 10

Pt

hc

(b) [Note: From the value of the photoelectric current, we can deduce how many electrons are

emitted per second.]Number of photoelectrons emitted per second

=

=

=

9

19

11 1

current

charge of each electron

48 10

1.60 10

3.0 10 s

Fraction of photons that produces electrons

=

=

=

11 1

18 1

7

number of electrons emitted per second

number of photons incident per second

3 10 s

3 10 s

1.0 10


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3(i) Power entering an area= intensity area

= (1.5 1011

) (3.5 103

)2

= 5.8 1016

W

(ii) Number of photons entering per second energy of photons entering per second

=

=

=

=

16

34 8 9

1

energy of photons entering per secondenergy of each photon

5.8 10

(6.63 10 )(3 10 ) / 550 10 )

1600 s

P

hc

[Note: the unit is s1

because it is number of photons enteringper second.]

4(a) Power of radiation incident on the silver surface

= intensity area

= (210) (12 106

)

= 2.52 103

W

Number of photons incident per second

=

=

=

=

3

34 8 9

15 1

energy of incident photons per second


2.52 10

(6.63 10 )(3 10 ) / 254 10 )

3.22 10 s

P

hc

(b) Rate of emission of electrons current

=

=

=

10

19

9 1

current

charge of each electron

4.8 101.60 10

3.0 10 s

(c)(i) Photoelectric quantum yield

=

=

=

9

15

7

number of electrons emitted per second

number of photons incident per second

3 10

3.22 10

9.3 10

(ii) 1. After absorbing a photon, an electron may collide with other particles and lose energy,


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resulting in insufficient energy to escape from the metal.

2. In order to escape, some electrons require more energy than what the photons can

provide.

(d) At longer wavelength, the frequency is lower, when the frequency is lower than the thresholdfrequency, the photon energy is less than the work function and so the photon is unable toeject any electron from the metal.

5 When one electron hits the 1sttarget, 5 electrons are ejected. These 5 electrons go on and

hit the 2ndtarget, 25 or 52

electrons are produced.

After the 9thtarget, 59

electrons are produced.

So for each electron that is ejected from the cathode, 59

electrons are produced at the end ofthe 9thtarget.

(i) Number of electrons emitted from the final target

=

=

=

3

19

14 1

current

charge of each electron0.10 10

1.60 10

6.25 10 s

current

Since 59

electrons are emitted from the final target for each photoelectron ejected from thecathode, number of photoelectrons ejected per second from the cathode

=

=

14

9

8 1

6.25 10

5

3.2 10 s

(ii) One in three of the incident photons succeeds in ejecting a photoelectron. This means that

for each photoelectron ejected there are 3 photons. So, number of photons incident per

second = 3 3.2 108

= 9.6 108

s1

Power of the light = total energy of photons incident per second

= number of photons incident per second energy of each photon

=

=

=

8

34 88

9

10

= number of photons incident per second energy of each photon

9.6 10

(6.63 10 )(3 10 )9.6 10

400 10

4.8 10 W

hc

_Photoelectric Effect_

6(a) Doubled. [Note: photon energy

19

31

2(2.46 10 )

9.11 10 ]

(b) Unchanged. [Note: work function depends on the type of metal.]

(c) Greater. [Note: KEmax= photon energy work function. Wrong to say that KEmaxis doubled]


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7 KEmax= photon energy work functioneVS= hf VS= (h/e)f /eSo for a graph of VSagainst f, the gradient is h/e.Answer: D.

8 KEmax= photon energy work functioneVS= hf KEmaxis dependent on fand Answer: D.

9 Gradient is h/ewhich is a constant, so both graphs should have the same gradient, soanswer is either C or D.The f-intercept (horizontal intercept) gives fmin, the threshold frequency, which directlydepends on the work function. A metal of greater work function means higher thresholdfrequency.Answer: D.

10 KEmax= photon energy work functioneVS= hf

KE does not depend on intensity]

Answer: D.

11(a)(i) When V is small, there are stray photoelectrons that dont reach the collector.When V is increased, some of these stray photoelectrons get attracted to the collectorand thus the photoelectric current increases.

With further increase in Vand when all the stray electrons reach the collector, the currentcannot increase any more as it is limited by the rate of emission of electrons from theemitter.

The rate of emission of electrons can only be increased if the intensity of light isincreased.

(ii) The electrons are emitted with different kinetic energies.Those electrons with high enough kinetic energy are still able to reach the gauze despitenegative V.

(b)(i) When intensity is increased, more photons are incident on the zinc plate, and so more

photoelectrons are emitted. With more photoelectrons reaching the gauze, the maximumphotoelectric current is increased.

(ii) The intensity of illumination does not affect the maximum kinetic energy of thephotoelectrons, this is because the maximum kinetic energy depends on the frequencyof the radiation which remains the same.

With the maximum kinetic energy unchanged, the stopping potential (required to stop the

most energetic electrons) is unchanged, so the value of V necessary to prevent any

photoelectric current (i.e. the stopping potential) remains constant.

12(a) Threshold frequency fmin= /h= 1.66 10-19/6.63 10-34= 2.50 1014Hz


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(b) KEmax = hf = (6.63 10-34)(6.22 1014) (1.66 10-19)= 2.46 10-19J

Convert to eV: 2.46 10-19J = 1.54 eV

(c) KEmax= mvmax2

vmax=

19

31

2(2.46 10 )

9.11 10

= 7.35 105m s-1

(d) KEmax=eVS

=

19

19

2.46 10

1.6 10S

V [1 eV = 1.6 10-19J]

= 1.54 eV

13(a) True. No photoelectric emission if wavelength longer than threshold wavelength(corresponds to frequency smaller than the threshold frequency).

(b) True. As retarding potential is increased, more electrons get stopped.

(c) True.

14(a) =

hc

=

=

34 8

9

19

(6.63 10 )(3 10 )310 10

6.42 10 J

(b) KEmax= photon energy work function= hf

=

=

34 819

9

19

(6.63 10 )(3 10 )6.42 10

240 10

1.87 10 J

Stopping potential VS = KEmax/e

=

=

19

19

1.87 10

1.6 10

1.17 V

15 Stopping potential VS= 2.0 V

(a) KEmax= eVS

= 1.60 10-19

2.0 = 3.20 10-19 J

= hf KEmax


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=

=

=

=

34 819

9

19

19

19

(6.63 10 )(3 10 )3.2 10

275 10

4.03 10 J

4.03 10eV

1.6 10

2.54 eV

(b) KEmax= mvmax2

vmax=

19

31

2(3.2 10 )

9.11 10

= 8.4 105m s-1

16 Note: intensity Iphotoelectric current iBut intensity I(wave amplitudeA)2

=

=

=

=

2

2

4

4

4

2

P

Q

P

Q

P

Q

P

Q

i

iI

I

A

A

A

A

17 eVS= hf Substituting the 2 set of values into the equation:e(1.22) = h(0.75 1015) - ------- (1)e(0.19) = h(0.50 1015) - ------- (2)

(a) (1) (2): 1.03e= (0.25 1015)hh= 1.03(1.6 10-19)/(0.25 1015)h= 6.59 10-34J s

(b) Sub value of hinto equation (2):= (6.59 10-34)(0.50 1015) - e(0.19)= 2.99 10-19J

_Wave Particle Duality_

18 KEmax= hf & f = c/Substituting the 2 set of values into the equation:

9.9 10-20= h(3 108)/(492 10-9) - ------- (1)3.8 10-20= h(3 108)/(579 10-9) - ------- (2)

(a) (1) (2): 6.1 10-20= h(3 108)(1/(492 10-9) 1/(579 10-9)]h= 6.66 10-34J s

(b) Sub value of hinto equation (1):= (6.66 10-34)(3 108)/(492 10-9) - 9.9 10-20= 3.07 10-19J


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(c) hfmin=fmin= /hfmin= (3.07 10

-19)/( 6.66 10-34)fmin= 4.61 10

14Hz

19(a) de Broglies wavelength of electron = h/pp= h/= 6.63 10-34/(0.15 10-9)p= 4.4 10-24kg m s-1

(b) KE E= mev2/2 =p2/2me

(c) mev2= eVp.d. [gain in KE = loss of E.P.E.]

p2/2me= eVp.d.. . . [me= 9.11 10

-31kg, e = 1.6 10-19 C]Vp.d.= 67 V

20 Electrons have same pattern as X-rays, so their de Broglie wavelength must = 0.10 nm

Usingp= h/, momentum of electron must be = h/(0.1 10-9)

Usingp2/2me= eVp.d.from Q19, Vp.d.= 151 V

21(a) de Broglie wavelength of a particle = h/p

KE E= mv2/2 =p2/2m

p=

= h/ mE2

(b) = h/ mE2

=)101()1011.9(2

1063.6

1831

34

= 5 10-10m

(c) This value is very close to the typical diameter of an atom. [Which makes possible theobservation of electron diffraction pattern when a beam of electron is directed at a crystal]

22 Answer: C

23(a) - Recall photon momentum = h/ (de Broglies)

- When photon hits the surface, it rebounds with equal momentum but in opposite direction

- momentum change = 2h/

- Each second, nphotons hit the surface

- So total momentum change per second = 2nh/

- Force = 2nh/

(b) - If photons are absorbed, change in momentum in each collision is h/and the force nh/

mE2


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_Line Spectra_

24(a) False as photon has no mass (just need to memorise as a piece of fact, reason beyondsyllabus)

(b) False. Same reason as (a)

25 Answer: B

26 Answer: A

27 E1= E2+ E3hc/1= hc/2+ hc/31/1= 1/2+ 1/3Answer: D

28

29 Answer: D

30 See lecture notes. There are 2 ways gain energy through collision with other particles orabsorb a photon. Simple way to bring about collision is to heat up the gas. If photons areused to raised energy level, the photon energies must match the energy gaps between theenergy levels of the hydrogen atoms

31 - The end of the spectrum with emission lines more closely packed corresponds to the higherfrequency lines.- Hence X must be the lowest frequency emission of a series

32 Answer: D

33(i)

Energy

f1 f2 f3

f4 f5

highfrequency

f3

low

frequency

f2

f1

f5

f4

bring down foreasier comparisonwith other 3transitions

-0.496 10-19

J

-3.136 10-19

J

-14.72 10-

J

0AB

C

D


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(ii) hc/ = EA EC

= hc/(EA EC)

= (6.63 10-34)(3.00 108)/[0 - (-3.136 10-19)]

= 6.34 10-7m (634 nm)

(iii) Level B to Level C

(iv) UV (EADabout 5 times of EAC using E = hc/, AC5AD. i.e. ADabout 120 nm)

(b) The above figure shows discrete energy levels.Atoms will therefore absorb only specific amounts of energy to go from the ground level tohigher levels.So when white light is passed through the atoms, only photons of certain wavelength areabsorbed, resulting in a line absorption spectra.

34 The re-emitted photons will be randomly emitted in all directions,

hence very few of the re-emitted photons will go towards the screen thereby causingrelatively dark lines.

35 Photons emitted by the sodium lamp are mostly absorbed by the sodium in the sodiumchloride vapour, and later re-emitted in all directions.As a result, there is less light falling on the screen, leading to dark shadow.

36 The fact that the emission spectrum of a nucleus is also a line spectrum suggests that anucleus too has discrete energy levels. (i.e. certain allowed energy states)However, energy levels are much further apart resulting in emission of high energy photonsin gamma region

37(a)(i) Energy change = 4.9 eV= 7.84 x 10-19J [1]

(ii) Wavelength = hc/E= 254 nm [1]

This is in the UV region. [1]

(b)(i) Transition from Level 1 to 2, or 1 to 3.

(ii) For 1 to 2, KE left is 2.1 eV. [1]For 1 to 3, KE left is 0.3 eV. [1]

(c) No transitions. [1]No energy transition is exactly 7 eV. [1]

(d) 10.4 eV [1](to ionize means to remove the bound electron to infinity, which corresponds to 0 energy)

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2011 Quantum I Solutions