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Flashcards in Unit 2 Deck (83)
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1
Q

State what is meant by ground state

A

When electrons are in their lowest energy and most stable state

2
Q

State what is meant by work function.

A

minimum energy required

to remove electron from metal

3
Q

What is the work function of a metal

A

The minimum energy needed by an electron to escape from the metal surface

4
Q

What was De Broglies hypothesis (2)

A

De Broglies hypothesis was that wavelength = h/p = h/mv and that matter particles have a duel wave/particle nature because: electrons they can be deflected in an electric/magnetic field (particle) and when accelerated through gaps between atoms in a crystalline graphite target the diffraction pattern shows they can behave like waves.

5
Q

Particle and wave behaviour of light examples

A

Particle behaviour of light: photoelectric effect

Wave behaviour of light: refraction and diffraction

6
Q

Explain the emission spectra and state where it is obtained from

A

if a gas or vapour emits light due only to de-excitation between atomic energy levels then the light it produces can be displayed as an emission ‘line spectrum’. The lines consist of a few beams of coloured light (the rest is blackness) that are diffracted in different directions by the spectroscope (obtained in labs using helium).

7
Q

Explain the absorption spectrum and state where it is obtained from

A

if we send thermal radiation through a non-excited gas then the atoms in the gas will absorb photons in those parts of the spectrum that cause excitation – that light will be missing from the spectrum (this is the absorption spectrum) (obtained from distant starts).

8
Q

What phenomenon can be used to demonstrate the wave properties of electrons?

A

Electron diffraction / interference / superposition

9
Q

In which part of the electromagnetic spectrum are the photons?

A

Ultraviolet

10
Q

What is meant by an excited mercury atom?

A

An electron (in ground state ) has moved in to higher energy level

11
Q

A fluorescent tube is filled with mercury vapour at low pressure. After mercury atoms have been excited they emit photons. How do the mercury atoms in the fluorescent tube become excited?

A

Free electrons collide with orbital electrons in atom transferring energy

12
Q

A fluorescent tube is filled with mercury vapour at low pressure. After mercury atoms have been excited they emit photons.
Why do the excited mercury atoms emit photons of characteristic frequencies?

A

(mercury) atoms have specific energy levels when electrons change levels they lose an specific amount of energy (photons emitted with specific amount of energy) (leading to photons of) specific frequencies

13
Q

Calculate the wavelength of electrons travelling at a speed of 2.5 × 105 m s–1.

A

(use of λ=h/mv)λ=6.63 × 10-34/(9.11 × 10-31×2.5×105)

λ=2.9 × 10-9m

14
Q

Explain how the coating on the inside of a fluorescent tube emits visible light.

A

Coating absorbs photons/uv light and re-emits photons of lower energy, longer wavelength and lower frequency

15
Q

Equation for energy in photon and wavelength

A

E = hf, f=c/lambda

16
Q

Equation for threshold frequency

A

E = phi/h

17
Q

Equation for Ekmax

A

Ekmax = hf - phi

18
Q

Explain why the kinetic energy of the emitted electrons has a maximum value

A

hf is a set energy available which has equal energy to photons - the energy required to remove the electron varies

19
Q

Explain with reference to the work function why, if the frequency of the radiation is
below a certain value, electrons are not emitted.

A
(work function is the) minimum energy needed to release
an electron (1)

below a certain frequency the energy of photon is less than
work function

20
Q

State a unit for work function.

A

Joule

21
Q

Light energy is incident on each square millimetre of the surface at a rate of
3.0 × 10–10 J s–1
. The frequency of the light is 1.5 × 1015 Hz.
(i) Calculate the energy of an incident photon.

A

(use of E = hf)

energy = 6.63 × 10–34 × 1.5 × 1015

energy = 9.9 × 10–19 (J) (1)

22
Q

Calculate the number of photons incident per second on each square millimetre of
the metal surface.

A

number of photons per second = 3.0 × 10–10/9.9 × 10–19

number of photons per second = 3.0 × 10^8

23
Q

In the wave theory model of light, electrons on the surface of a metal absorb energy from a small area of the surface.

(i) The light striking the surface delivers energy to this small area at a rate of
3. 0 × 10–22 J s–1

The minimum energy required to liberate the electron is 6.8 × 10–19 J.
Calculate the minimum time it would take an electron to absorb this amount of
energy.

A

(time taken = 6.8 × 10–19/3 × 10–22)

time taken = 2.3 × 10^3 s

24
Q

In practice the time delay calculated in part c (i) does not occur. Explain how this
experimental evidence was used to develop the particle model for the behaviour of
light.

A

light travels as particles/ photons (1)

(which transfer) energy in discrete packets

25
Q

When light of a certain frequency is shone on a particular metal surface, electrons are
emitted with a range of kinetic energies.
(a) Explain
• in terms of photons why electrons are released from the metal surface, and
• why the kinetic energy of the emitted electrons varies upto a maximum value.
The quality of your written communication will be assessed in this question.

A

energy is needed to remove an electron from the surface
work function φ (of the metal) is the minimum energy needed by
an electron to escape from the surface
light consists of photons , each of energy E = hf

ONE PHOTON IS ABSORBED BY ONE ELECTRON

an electron can escape (from the surface) if hf > φ
kinetic energy of an emitted electron cannot be greater than hf – φ
an electron below the surface needs to do work/uses energy to reach
the surface
kinetic energy of such an electron will be less than hf – φ

26
Q

The graph below shows how the maximum kinetic energy of the electrons varies with the
frequency of the light shining on the metal surface.

What is the threshold frequency, where would a metal with higher threshold frequency be, what is the gradient, and what are the axis’

A

threshold frequency - x axis
parallel line to the right = higher threshold frequency

gradient = Planck’s constant

27
Q

The threshold frequency of a particular metal surface is 5.6 × 1014 Hz. Calculate the
maximum kinetic energy of emitted electrons if the frequency of the light striking the metal
surface is double the threshold frequency.

A

(use of hf = phi )
hf = 6.63 × 10–34 × 2 × 5.6 × 1014
phi = 3.7(1) × 10–19 J
Ek= 2 × 3.7 × 10–19 – 3.7 × 10–19 = 3.7 × 10^–19 J

28
Q

The photoelectric effect suggests that electromagnetic waves can exhibit particle-like
behaviour. Explain what is meant by threshold frequency and why the existence of a
threshold frequency supports the particle nature of electromagnetic waves.
The quality of your written communication will be assessed in this question.

A

threshold frequency minimum frequency for emission of electrons
• if frequency below the threshold frequency, no emission
even if intensity increased
• because the energy of the photon is less than the work function
• wave theory can not explain this as energy of wave
increases with intensity
• light travels as photons
• photons have energy that depends on frequency
• if frequency is above threshold photon have enough energy
• mention of lack of time delay

29
Q

An alpha particle of mass 6.6 × 10–27 kg has a kinetic energy of 9.6 × 10–13 J.
Show that the speed of the alpha particle is 1.7 × 107
m s–1

Calculate the momentum of the alpha particle, stating an appropriate unit.

A

use of Ek = ½ v^2
½× 6.6 × 10–27 × v^2 = 9.6 × 10–13
v^2 = 2.91 × 10–14
v = 1.7 × 10^7m/s

(use of p = mv)
p = 6.6 × 10–27 × 1.7 × 107
(1)
p = 1.1 × 10–19 (1) kg m s–1
/N s
30
Q

Calculate the de Broglie wavelength of the alpha particle.

A

(use of λ= )
λ= 6.63 × 10–34/1.1 × 10–19 (1)
λ= 5.9 × 10–15 m (1) (6.03 × 10–15 m)

31
Q

When monochromatic light is shone on a clean metal surface, electrons are emitted from
the surface due to the photoelectric effect.
(a) State and explain the effect on the emitted electrons of
(i) increasing the frequency and (ii) intensity of the light,

A

(i) the (maximum) kinetic energy/speed/velocity/momentum
of released electrons increases (1)
this is because increasing the frequency of the photons increases
their energy or correct application of photoelectric equation (1)
(ii) the number of electrons emitted (per second) increases (1)
because there are now more photons striking the metal surface
(per second)

32
Q

The wave model was once an accepted explanation for the nature of light. It was rejected
when validated evidence was used to support a particle model of the nature of light.
Explain what is meant by validated evidence.

A

experiment/observation needs to be performed (to test a theory) (1)
the results of (the experiment) need to be proved/repeatable/replicated/
confirmed (1)
[or threshold frequency (1) could not be explained by the wave
model (1)]

33
Q

The threshold frequency of lithium is 5.5 × 1014 Hz.
(i) Calculate the work function of lithium, stating an appropriate unit,

(ii) Calculate the maximum kinetic energy of the emitted electrons when light of
frequency 6.2 × 1014 HZ is incident on the surface of a sample of lithium.

A

(use of phi= hf)
phi = 6.63 × 10–34 × 5.5 × 10^14
phi = 3.65 × 10^-19J

Ek = 6.63 × 10–34 × 6.2 × 1014 (1) – 3.65 × 10–19 (1)
Ek = 4.6 × 10–20 J (accept 5.1 × 10–20 J
34
Q

Explain why the emitted electrons have a range of kinetic energies up to a maximum
value.

A

photons have energy dependent on frequency OR energy of photons constant
one to one interaction between photon and electron
Max KE = photon energy – work function in words or symbols
more energy required to remove deeper electrons

35
Q
When monochromatic light is shone on a clean cadmium surface, electrons with a
range of kinetic energies up to a maximum of 3.51 × 10–20 J are released. The work
function of cadmium is 4.07 eV. Calculate the frequency of the light
A

(use of hf = Ø + Ek(max)
)
6.63 × 10–34 × f = 4.07 × 1.60 × 10–19 + 3.51 × 10–20
f = 1.04 × 1015 (Hz) OR 1.03 × 1015 (Hz) (3 sig figs)

36
Q

In order to explain the photoelectric effect the wave model of electromagnetic radiation was
replaced by the photon model. Explain what must happen in order for an existing scientific
theory to be modified or replaced with a new theory.

A

theory makes predictions tested by repeatable/checked by other
scientists/peer reviewed (experiments) OR new evidence that is repeatable/
checked by other scientists/peer reviewed

37
Q

Energy level 1 = 13.6
Energy level 2 = 3.4
Energy level 3 = 1.51
Energy level 4= 0.85

Show that the frequency of spectral line B is about 4.6x10^14Hz

Calculate the energy, in eV, of the longest wavelength of electromagnetic radiation
emitted during this process.

A
(3.40-1.51 = 1.89)
ΔE= 1.89 × 1.60 × 10–19(J) (1)
(= 3.02 × 10–19(J))

f(ΔE/h) = 4.56x10^14Hz

(1.51–0.85) = 0.66(eV

38
Q

The hydrogen atom is excited and its electron moves to level 4.
(i) How many different wavelengths of electromagnetic radiation may be emitted as the
atom returns to its ground state?

A

6

39
Q

In a fluorescent tube, explain how the mercury vapour and the coating of its inner surface
contribute to the production of visible light. You may be awarded additional marks to those
shown in brackets for the quality of written communication in your answer.

A

mercury vapour at low pressure is conducting (1)
atoms of mercury are excited by electron impact (1)
producing (mainly) ultra violet radiation (1)
which is absorbed/ excites the coating (1)
which, upon relaxing, produces visible light (1)
electrons cascade down energy levels (1)

41
Q

The diagram below shows part of an energy level diagram for a hydrogen atom.
n = 4 –0.85 eV
n = 3 –1.50 eV
n = 2 –3.40 eV
n = 1 –13.60 eV
(a) The level, n = 1, is the ground state of the atom.
State the ionisation energy of the atom in eV.
Calculate the wavelength of the photon with the smallest energy.

A

ionisation energy = 13.6eV

energy in Joules = 1.90 (1) × 1.6 × 10–19 = 3.04 × 10–19 (J) (1)
(use of E = hc/λ)
3.04 × 10–19 = 6.63 × 10–34 × 3 × 108
/λ(1)
(working/equation must be shown)
λ= 6.54 × 10^–7m
42
Q

State what is meant by the ionisation of an atom.

A

when an atom loses an orbiting electron (and becomes charged)

43
Q

Explain the similarity and difference between excitation and ionisation.

A
in either case an electron receives (exactly the right amount of) energy
excitation promotes an (orbital) electron to a higher energy/up a level
ionisation occurs (when an electron receives enough energy) to leave the atom (ground state)
44
Q

The atom in the ground state is given 5.00 × 10–17 J of energy by electron impact.
(i) State what happens to this energy.
(ii) Describe and explain what could happen subsequently to the electrons in the higher
energy levels.

A

the electron in the ground state leaves the atom (1)
with remaining energy as kinetic energy (0.89 × 10−17 J) (1)

(ii) the orbiting electrons fall down (1)
to fill the vacancy in the lower levels (1)
various routes down are possible (1)
photons emitted (1)
taking away energy (1)

45
Q

An atom can also become excited by the absorption of photons. Explain why only photons
of certain frequencies cause excitation in a particular atom.

A

electrons occupy discrete energy levels
and need to absorb an exact amount of/enough energy to move to a higher level
photons need to have certain frequency to provide this energy or e = hf
energy required is the same for a particular atom or have different energy levels
all energy of photon absorbed
in 1 to 1 interaction or clear a/the photon and an/the electrons

46
Q

The ionisation energy of hydrogen is 13.6 eV. Calculate the minimum frequency necessary
for a photon to cause the ionisation of a hydrogen atom. Give your answer to an
appropriate number of significant figures.

A
energy = 13.6 × 1.60 × 10−19 = 2.176 × 10−18 (J)
hf = 2.176 × 10−18

f = 2.176 × 10−18 ÷ 6.63 × 10−34 = 3.28 × 1015 Hz

47
Q

What is meant by an excited atom?

A

an electron/atom is at a higher level than the ground state (1)
or electron jumped/moved up to another/higher level

48
Q

Describe the process by which mercury atoms become excited in a fluorescent tube

A

electrons (or electric current) flow through the tube (1)
and collide with orbiting/atomic electrons or mercury atoms (1)
raising the electrons to a higher level (in the mercury atoms)

49
Q

What is the purpose of the coating on the inside surface of the glass in a fluorescent
tube?

A

photons emitted from mercury atoms are in the ultra
violet (spectrum) or high energy photons (1)
these photons are absorbed by the powder or powder changes
frequency/wavelength (1)
and the powder emits photons in the visible spectrum (1)
incident photons have a variety of different wavelengths

50
Q

n=4 = -0.26x10^-18J
n=3 = -0.59x10^-28J
Calculate the frequency of an emitted photon due to the transition level n = 4 to level
n = 3.

A

(use of E = hf)
–0.26 × 10–18 – 0.59 × 10–18 (1) = 6.63 × 10–34 × f (1)
f = 0.33 × 10–18/(6.63 × 10–34) = 5.0 × 1014 (Hz)

51
Q

A proton and an electron have the same velocity. The de Boglie wavelength of the electron
is 3.2 × 10–8
m.
(a) Calculate,
(i) the velocity of the electron,
(ii) the de Broglie wavelength of the proton

A

i) lambda = h/mv
v=
6.63x10^-34/9.1x10^-31 x 3.2x10^-8
= 2.3x10^-4m/s

ii) lambda = h/mv =
h/ 1.67x10^-27 x 2.27x10^4 =
1.7x10^-11m

52
Q

State why it is easier to demonstrate the wave properties of electrons than to
demonstrate wave properties of protons

A

easier to obtain electrons

53
Q

State what kind of experiment would confirm that electrons have a wave-like nature.
Experimental details are not required.

A

Diffraction

54
Q

When can the photoelectric emission of electrons from a metals surface not occur

A

When the frequency of the incident electromagnetic radiation is below the threshold frequency.

55
Q

What must be true for photoemission to occur

A

hf > phi

56
Q

1 (mega)/electron volt = how many joules

A
  1. 6x10^-19

1. 6x10^-13

57
Q

What is the ionisation energy

A

the minimum energy to remove an electron from an atom from the ground state

58
Q

Maximum kinetic energy =

A

hf - phi
OR
electron charge x Voltage (Voltage in eV)

59
Q

One similarity and one difference between ionisation and excitation

A

Similarity: energy is absorbed by the atom
Difference: electron stays in atom when excitation occurs but leaves the atom when ionisation occurs

60
Q

What is an electron volt

A

The work done when an electron moves through a potential difference of one volt

61
Q

Work done =

A

charge x voltage

62
Q

What is the effect of changing the wavelength on kinetic energy

A

Nothing

63
Q

Fluorescent light bulbs 2 pros

A

energy consumption than filament bulbs, last longer than incandescent lamps

64
Q

Fluorescent light bulbs 2 cons

A

not environmentally friendly, mercury is highly toxic, disposal has to be carried carefully as could toxicate water supply.

65
Q

When is a photon emitted

A

when an atom de-excites due to an electron moving to an inner shell

66
Q

Kinetic energy =

A

0.5mv^2

67
Q

The ______ the gap, the greater the amount of diffraction

A

narrower

more curved at the end of semicircle

67
Q

Explain why, when ultraviolet light is shone on a positively charged plate, no charge is lost by the plate.

A

The process involves the ejection of electrons which are negatively charged.Any electrons ejected will only make the positive charge greater.

68
Q

Explain how light is emitted from mercury light

A

Electrons are thermally emitted from a heated cathode (thermionic emission) and the anode accelerates them so that they gain sufficient energy to excite mercury atoms. The electrons excite the mercury atoms and when they de-excite they emit ultraviolet rays which hit the phosphorous coating exciting it and when it de-excites it emits visible light rates.

69
Q

Describe and explain the experiment portraying the photoelectric effect

A

the gold foil drops to zero when the zinc plate has negative charge and illuminated by uv light because conduction electrons at the zinc surface leave the zinc surface when the uv light is directed at it. The emitted electrons are photoelectrons which take negative charge with them. If the zinc plates positively charged then the leaf rises and stays in position regardless of the uv light. If the electroscope is charged negatively or positively and visible light is directed at it the lead rises and stays in position.

70
Q

What causes ionisation

A

When alpha, beta and gamma radiation colliding with atoms which knock an orbital electron out of the atom leaving a positively charged ion or electrons passing through a fluorescent tube creating ions when they collide with the atoms of the gas in the tube

71
Q

When can ionisation not occur

A

if kinetic energy is less than ionisation energy ionisation cannot occur

72
Q

Excitation occurs when a…

A

orbital electron gains energy and moves to a higher energy state (electrons only absorb specific energies)

73
Q

What did wave theory incorrectly predict (2)

A

That emission should take place with waves of any frequency and that emission would take longer using low intensity waves than using high intensity waves.

74
Q

What did Einstein say to explain photoelectricity (2)

A

When light is incident on a metal surface, an electron at the surface absorbs a single photon from the incident light and therefore gains energy equal to hf. An electron can leave the metal surface if the energy gained from a single photon exceeds the work function of the metal.

75
Q

De Broglies equation

A

Lambda = h/p = h/mv

76
Q

Unit for momentum

A

kgm/s

77
Q

Unit for kinetic energy, velocity and threshold frequency

A

Joules, m/s, Hz

78
Q

1 nanometre = ____metres

A

1x10^-9m

79
Q

Emission spectrum

A

All black but few lines where an electron is moving from a higher energy level to a lower energy level and emitting a photon of light of energy equal to the difference in the energy of the two levels

80
Q

Energy of photon emitted in emission spectra when an electron moves from one energy level to another and emits a photon, hf =

A

E1 - E2

81
Q

Absorption spectra

A

When white light passes through a gas the gas absorbs particular wavelengths of light causing a few dark lines in an almost continuous spectra

82
Q

The graph of frequency against Ekmax shows what

A

The gradient is h
The x intercept is fmin = phi/h
The y intercept is the the negative of phi the threshold frequency

83
Q

Photoelectric effect =

A

photons of light incident on metal surface cause the emission of electrons near the metal’s surface