Unit 2

Question 1

State what is meant by ground state

Answer 1

When electrons are in their lowest energy and most stable state

Question 2

State what is meant by work function.

Answer 2

minimum energy required

to remove electron from metal

Question 3

What is the work function of a metal

Answer 3

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

Question 4

What was De Broglies hypothesis (2)

Answer 4

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.

Question 5

Particle and wave behaviour of light examples

Answer 5

Particle behaviour of light: photoelectric effect

Wave behaviour of light: refraction and diffraction

Question 6

Explain the emission spectra and state where it is obtained from

Answer 6

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).

Question 7

Explain the absorption spectrum and state where it is obtained from

Answer 7

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).

Question 8

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

Answer 8

Electron diffraction / interference / superposition

Question 9

In which part of the electromagnetic spectrum are the photons?

Answer 9

Ultraviolet

Question 10

What is meant by an excited mercury atom?

Answer 10

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

Question 11

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?

Answer 11

Free electrons collide with orbital electrons in atom transferring energy

Question 12

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?

Answer 12

(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

Question 13

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

Answer 13

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

λ=2.9 × 10-9m

Question 14

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

Answer 14

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

Question 15

Equation for energy in photon and wavelength

Answer 15

E = hf, f=c/lambda

Question 16

Equation for threshold frequency

Answer 16

E = phi/h

Question 17

Equation for Ekmax

Answer 17

Ekmax = hf - phi

Question 18

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

Answer 18

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

Question 19

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

Answer 19

(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

Question 20

State a unit for work function.

Answer 20

Joule

Question 21

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.

Answer 21

(use of E = hf)

energy = 6.63 × 10–34 × 1.5 × 1015

energy = 9.9 × 10–19 (J) (1)

Question 22

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

Answer 22

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

number of photons per second = 3.0 × 10^8

Question 23

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.

Answer 23

(time taken = 6.8 × 10–19/3 × 10–22)

time taken = 2.3 × 10^3 s

Question 24

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.

Answer 24

light travels as particles/ photons (1)

(which transfer) energy in discrete packets

Question 25

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.

Answer 25

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

Question 26

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’

Answer 26

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

gradient = Planck’s constant

Question 27

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.

Answer 27

(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

Question 28

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.

Answer 28

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

Question 29

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.

Answer 29

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

Question 30

Calculate the de Broglie wavelength of the alpha particle.

Answer 30

(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)

Question 31

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,

Answer 31

(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)

Question 32

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.

Answer 32

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)]

Question 33

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.

Answer 33

(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

Question 34

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

Answer 34

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

Question 35

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

Answer 35

(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)

Question 36

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.

Answer 36

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

Question 37

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.

Answer 37

(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

Question 38

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?

Answer 38

6

Question 39

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.

Answer 39

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)

Question 41

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.

Answer 41

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

Question 42

State what is meant by the ionisation of an atom.

Answer 42

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

Question 43

Explain the similarity and difference between excitation and ionisation.

Answer 43

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)

Question 44

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.

Answer 44

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)

Question 45

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

Answer 45

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

Question 46

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.

Answer 46

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

Question 47

What is meant by an excited atom?

Answer 47

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

Question 48

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

Answer 48

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)

Question 49

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

Answer 49

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

Question 50

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.

Answer 50

(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)

Question 51

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

Answer 51

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

Question 52

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

Answer 52

easier to obtain electrons

Question 53

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

Answer 53

Diffraction

Question 54

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

Answer 54

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

Question 55

What must be true for photoemission to occur

Answer 55

hf > phi

Question 56

1 (mega)/electron volt = how many joules

Answer 56

1. 6x10^-19

1. 6x10^-13

Question 57

What is the ionisation energy

Answer 57

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

Question 58

Maximum kinetic energy =

Answer 58

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

Question 59

One similarity and one difference between ionisation and excitation

Answer 59

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

Question 60

What is an electron volt

Answer 60

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

Question 61

Work done =

Answer 61

charge x voltage

Question 62

What is the effect of changing the wavelength on kinetic energy

Answer 62

Nothing

Question 63

Fluorescent light bulbs 2 pros

Answer 63

energy consumption than filament bulbs, last longer than incandescent lamps

Question 64

Fluorescent light bulbs 2 cons

Answer 64

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

Question 65

When is a photon emitted

Answer 65

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

Question 66

Kinetic energy =

Answer 66

0.5mv^2

Question 67

The ______ the gap, the greater the amount of diffraction

Answer 67

narrower

more curved at the end of semicircle

Question 67

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

Answer 67

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

Question 68

Explain how light is emitted from mercury light

Answer 68

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.

Question 69

Describe and explain the experiment portraying the photoelectric effect

Answer 69

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.

Question 70

What causes ionisation

Answer 70

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

Question 71

When can ionisation not occur

Answer 71

if kinetic energy is less than ionisation energy ionisation cannot occur

Question 72

Excitation occurs when a…

Answer 72

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

Question 73

What did wave theory incorrectly predict (2)

Answer 73

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.

Question 74

What did Einstein say to explain photoelectricity (2)

Answer 74

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.

Question 75

De Broglies equation

Answer 75

Lambda = h/p = h/mv

Question 76

Unit for momentum

Answer 76

kgm/s

Question 77

Unit for kinetic energy, velocity and threshold frequency

Answer 77

Joules, m/s, Hz

Question 78

1 nanometre = ____metres

Answer 78

1x10^-9m

Question 79

Emission spectrum

Answer 79

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

Question 80

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

Answer 80

E1 - E2

Question 81

Absorption spectra

Answer 81

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

Question 82

The graph of frequency against Ekmax shows what

Answer 82

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

Question 83

Photoelectric effect =

Answer 83

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