Download Quantum Physics

8866 H1 Physics – J2/2011
11.Quantum Physics
Quantum Physics
Content
1.
2.
3.
4.
5.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
Energy of a photon
The Photoelectric effect
Wave-particle duality
Energy levels in atoms
Line spectra
Show appreciation of the particulate nature of electromagnetic radiation.
Recall and use E = hf.
Show an understanding that the photoelectric effect provides evidence for a
particulate nature of electromagnetic radiation while phenomena such as interference
and diffraction provide evidence for a wave nature.
Recall the significance of threshold frequency.
2
Recall and use the equation 21 mv max
= eVs, where Vs is the stopping potential.
Explain photoelectric phenomena in terms of photon energy and work function
energy.
Explain why the maximum photoelectric energy is independent of intensity whereas
the photoelectric current is proportional to intensity.
2
Recall, use and explain the significance of hf = φ + 21 mv max
.
Describe and interpret qualitatively the evidence provided by electron diffraction for
the wave nature of particles.
h
Recall and use the relation for the de Broglie wavelength λ = .
p
Show an understanding of the existence of discrete electron energy levels in isolated
atoms (e.g. atomic hydrogen) and deduce how this leads to spectral lines.
Distinguish between emission and absorption line spectra.
Recall and solve problems using the relation hf = E1 − E2.
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8866 H1 Physics – J2/2011
(a)
11.Quantum Physics
Show an appreciation of the particulate nature of electromagnetic radiation
Electromagnetic energy is emitted, transmitted and absorbed in discrete packets or
quanta. A quantum of electromagnetic radiation is known as a photon.
Eg.1:
(Ans: E)
Photon is the name given to
A
B
C
D
E
(b)
an electron emitted from a metal surface by the action of light.
a unit of energy.
a positively charged atomic particle.
an electron emitted from a metal surface by the action of heat.
a quantum of electromagnetic radiation.
Recall and use E = hf
The energy of a photon, E, is proportional to its frequency f, and is expressed as
E = hf = h
c
λ
where h is the Planck constant (= 6.63×10-34 Js).
Eg.2: Find the energy of a photon of electromagnetic radiation of wavelength 0.15 nm. In
which region of the electromagnetic spectrum is this radiation?
Answer:
E = hf = h
c
(3 × 108 ms-1 )
= (6.63×10-34 Js)
= 1.33×10-15 J.
λ
(0.15 × 10−9 m)
X-ray.
Recall:
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Eg.3:
11.Quantum Physics
(Ans: D)
The wavelength of a 5 MeV γ-ray is
A
B
C
D
E
4.95×10-38 m
8.89×10-32 m
8.89×10-30 m
2.49×10-13 m
2.48×10-10 m
Answer:
Data: 1 MeV = (1.6×10-19) x (1000 000) = 1.6×10-13 J
E = 5 MeV = 5 (1.6×10-13 J) = 8.0×10-13 J
E = hf = h
Eg.4:
c
λ
Î
λ=
hc (6.6 × 10−34 Js)(3 × 108 ms-1 )
=
= 2.49×10-13 m.
E
(8.0 × 10−13 J)
(Ans: D)
A photon of light enters a block of glass after traveling through a vacuum. The
energy of the photon on entering the glass block
A
B
C
D
E
increases because its associated wavelength decreases.
decreases because the speed of the radiation decreases.
stays the same because the speed of the radiation and the associated
wavelength do not change.
stays the same because the frequency of the radiation does not change.
stays the same because the speed of the radiation and its wavelength
increase by the same order.
Answer:
Only speed and wavelength change, frequency is unchanged.
(d)
Recall the significance of threshold frequency
(f)
Explain photoelectric phenomena in terms of photon energy and work function
energy.
(g)
Explain why the maximum photoelectric energy is independent of intensity
whereas the photoelectric current is proportional to intensity
Photoelectric phenomena:
Generally, electrons may be emitted from a metal surface by 3 methods:
(1)
by friction, e.g. by rubbing a metal surface vigorously with a piece of cloth,
(2)
by thermionic emission, i.e. by heating a metal such that the electrons in it
would gain enough kinetic energy to overcome the attraction of the metal
surface and escape from the surface (almost as though they are boiled off the
surface).
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(3)
11.Quantum Physics
by photoelectric effect, i.e. to incident a beam of light (em radiation) on a
metal surface such that the electrons in it would gain enough kinetic energy
from the incident light to overcome the attraction of the metal surface and
escape from the surface.
Photoelectric effect is the emission of electrons from the metal
surface when an electromagnetic radiation of high enough frequency
is incident on the metal.
Experimental Setup:
incident radiation
mA
C
A
V
Procedure:
In the above setup, radiation (or light) of known frequency f and intensity I is incident
onto an emitting electrode (cathode C) placed within an evacuated glass envelope,
together with the collecting electrode (anode A ). Electrons may be emitted by C and
collected by A. The electric potential of A may be varied and made positive or
negative (higher or lower) with respect to C.
Refer to Annex A for results and explanation.
(c)
Show an understanding that the photoelectric effect provides evidence for a
particulate nature of electromagnetic radiation while phenomena such as
interference and diffraction provide evidence for a wave nature
It is well-known that only waves are able to exhibit phenomena such as
interference and diffraction. Light as an example of an em radiation does exhibit
such phenomena, and hence may be considered as a wave.
On the other hand, the results of the photoelectric effect showed the inadequacy of
the wave nature of light. By considering light as a wave, only the first result can be
explained (i.e. photoelectric current ∝ intensity).
All the experimental results can be explained adequately by considering light as a
stream of photons. A photon is a quantum of em radiation, which may be
considered as a particle without mass, just a packet of energy.
*
[The explanation for the experimental results of the photoelectric effect was proposed
by none other than Albert Einstein. He was most famous for his Theory of Relativity,
but he won his Nobel Prize for Physics due to this theory (photoelectric effect) which
assumed that light may be considered as a stream of photons. The equation that
summarises the photoelectric effect was named Einstein’s equation. ]
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Eg.5:
Eg.6:
11.Quantum Physics
(Ans: B)
Which one of the following statements, referring to photoelectric emission, is always
true?
A
No emission of electrons occurs for very low intensity.
Not true. As long as there are photons with enough energy, there will
be electrons emitted. It is independent of intensity.
B
For a given metal there is a minimum frequency of radiation below which no
emission occurs.
True. This is the threshold frequency for that metal.
C
The velocity of the emitted electrons is proportional to the intensity of the
incident radiation.
Not true. It is the photoelectric current that is proportional to intensity.
D
The number of electrons emitted per second is independent of the intensity of
the incident radiation.
Not true. Number of electrons emitted per second is a measure of the
photoelectric current, which is proportional to intensity.
E
The number of electrons emitted per second is proportional to the frequency
of the incident radiation.
Not true. Number of electrons emitted per second is a measure of the
photoelectric current, which is proportional to intensity, not frequency.
(Ans: E)
A source emits monochromatic light of wavelength λ at power P. Given that h is the
Planck constant and c the speed of light, the rate of emission of photons is
A
Pc
hλ
B
λc
Ph
C
hc
Pλ
D
Ph
cλ
E
Pλ
hc
Answer:
Each photon has energy, E = hf. If N is the number of photons emitted in time t, the
c
total energy emitted in time t can be expressed as NE = Nhf = Nh
λ
total energy emitted Nhf Nhc
The source emits light at a power,
P=
=
=
time taken
t
tλ
N Pλ
So, the rate of emission of photons,
=
t hc
Eg.7:
(Ans: A)
A beam of monochromatic radiation falls on to a metal X and photoelectrons are
emitted. The rate of emission of photoelectrons will be double if
A
B
C
D
E
a beam of double the intensity is used.
radiation of double the frequency is used.
radiation of double the wavelength is used.
the thermodynamic temperature of the metal is doubled.
a metal with a work function half that of X is substituted for X.
Answer:
Double the rate of emission of photoelectrons means double the photoelectric
current. Since photoelectric current ∝ intensity, the intensity must also be doubled.
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(h)
11.Quantum Physics
2
Recall, use and explain the significance of hf = φ + 21 mv max
The above equation is known as the Einstein’s photoelectric equation.
When an electron absorbs energy hf from an incident photon, spends a minimum
energy φ to be emitted from the metal surface, it will have a maximum kinetic energy,
1
2
Note: (i)
(ii)
(e)
2
mv max
= hf − φ
2
mv max
belongs to an electron, measured by the stopping potential,
Vs, of the electrode A, which is made negative (lower potential) with
respect to C. This represents the minimum negative potential required
to stop even the most energetic electron from reaching the electrode A
(indicated by the zero current in the milliammeter), therefore,
2
1
mv max
= eVs.
2
1
2
φ is a characteristic of the electrode C, (= hfo, where fo is the threshold
frequency). and hf is the energy of a photon.
Recall and use the equation
1
2
2
mv max
= eVs, where eVs is the stopping potential
This just means, for the photoelectron:
loss of kinetic energy = gain in electric potential energy
Einstein’s equation can hence be written as:
eVs = hf − φ
eVs = hf − hfo
h
h
Vs = ( )f − ( )fo
e
e
φ
h
Vs = ( )f −
e
e
or
or
or
Vs
slope =
−
Note: (i)
φ
e
fo
h
e
f
The graph of Vs versus f has a gradient of
h
which is always
e
constant.
(ii)
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Different material of C will give a different value of φ, and
hence the graph will have different y-intercept.
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Eg.8:
11.Quantum Physics
(Ans: E)
The result of an experiment to investigate the energy of photoelectrons emitted from
a metallic surface is represented by the figure below.
stopping
potential
0
The gradient of the graph depends on the
A
B
C
D
E
Eg.9:
frequency
intensity of the incident radiation.
wavelength of the incident radiation.
work function of the irradiated surface.
pressure of residual gas in contact with the surface.
ratio of the Planck constant to the electronic charge.
(Ans: B)
Light quanta each of energy 3.5×10-19 J fall on the cathode of a photocell. The
current through the cell is just reduced to zero by applying a reverse voltage to make
the cathode 0.25 V positive with respect to the anode. The minimum energy required
to remove an electron from the cathode is
A 2.9×10-19 J
B 3.1×10-19 J
C 3.5×10-19 J
D 3.9×10-19 J
E 6.4×10-19 J
Answer:
Data: hf = 3.5×10-19 J, Vs = 0.25 V.
Question asks for minimum energy required to remove an electron from the cathode,
i.e. the work function energy of the cathode, φ.
eVs = hf − φ
φ = hf − eVs
= (3.5×10-19 J) − (1.6×10-19 C)(0.25 V) = 3.1×10-19 J
Using
Î
Eg.10:
(Ans: A)
In a photoelectric experiment, electrons are ejected from metals X and Y by light of
frequency f. The potential difference V required to stop the electrons is measured for
various frequencies.
If Y has a greater work function than X, which graph illustrates the expected results?
V
V
A
X
X
Y
0
V
B
C
Y
V
X
Y
f 0
V
D
Y
E
Y
X
X
f 0
f 0
f 0
f
Answer:
h
), so answers are A or D.
e
If Y has a greater work function than X, then the graph for Y should have a more
negative y-intercept.
First, the gradient of the graph cannot change (always =
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(i)
11.Quantum Physics
Describe and interpret qualitatively the evidence provided by electron
diffraction for the wave nature of particles
When a beam of electron is passed through a thin film of crystal (e.g. graphite ), the
dispersion pattern of the emergent electrons produced on a screen ( coated with
fluorescent, (c) ) is observed to be similar to the diffraction pattern produced by a
beam X-ray (a well-known electromagnetic wave, (b) ). This is known as the electron
diffraction phenomenon.
(j)
This phenomenon provides evidence for the wave nature of particles like
electrons.
h
Recall and use the relation for the de Broglie wavelength λ =
p
de Broglie suggested that: for a particle of momentum p (= mv), which exhibits wave
behaviour, it will have an associated wavelength λ, given by
λ=
h
p
Similarly, for electromagnetic radiation of wavelength λ, which exhibits particle
behaviour, it will have an associated momentum p, given by
h
p=
λ
This is known as the de Broglie principle.
Eg.11:
(Ans: D)
Light of frequency 5×1014 Hz consists of photons of momentum
A 4.0×10-40 kg m s-1
D 1.1×10-27 kg m s-1
B 3.7×10-36 kg m s-1
E 3.3×10-19 kg m s-1
C 1.7×10-28 kg m s-1
Answer:
h hf (6.63 × 10 −34 Js)(5 × 1014 Hz)
p= =
=
= 1.1×10-27 kg m s-1.
λ c
3.0 × 10 8 ms -1
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Eg.12:
11.Quantum Physics
(Ans: E)
A beam of light of wavelength λ is totally reflected at normal incidence by a plane
mirror. The intensity of the light is such that photons hit the mirror at a rate n. Given
that the Planck constant is h, the force exerted on the mirror by this beam is
A nhλ
B
nh
λ
C 2 nhλ
D
2nλ
h
E
2nh
λ
Answer:
h
λ
When totally reflected at normal incidence by a plane mirror,
h
the change in momentum = 2
λ
Since there are n photons hitting the mirror per unit time,
h
Δp
the rate of change of momentum,
= 2n
t
λ
h
Δp
By Newton’s 2nd law of motion, force, F =
= 2n
t
λ
Each photon has a momentum p =
(k)
Show an understanding of the existence of discrete energy levels in isolated
atoms (e.g. atomic hydrogen) and deduce how this leads to spectral lines
• The electrons orbiting an atom can only
occupy certain allowed orbits.
• In these allowed orbits, the total kinetic and
potential energies of all the electrons in the
atom will give rise to the specific energy
state of the atom e.g. E1, E2 and so on.
E4
• In the normal state or ground state of an
atom, its electrons will occupy the lowest
allowed orbit so that the energy of the atom
is at its lowest.
E3
E2
• Any change usually involves the movement
of the electrons to higher allowed orbits.
For the different allowed orbits occupied,
the atom will have different but specific
energy states, or excited states.
E1
• In this transition to a higher orbit, the energy
absorbed by the atom in excitation, the
excitation energy, is equal to the
difference between the energy levels of the
two states.
Energy level diagram of an atom having
various allowed states. The lowest energy
state E1 is the ground state. All others are
excited states.
• The highest excited state is when the
electron is at infinity (i.e. the atom is
ionised), so the highest excitation energy is
the ionisation energy.
Note the bigger gaps between energy
levels in the lower region (say, between
E2 and E1) than those in the higher
region (say, between E4 and E3).
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11.Quantum Physics
• On the other hand, if an electron makes a transition down from a higher excited
state to a lower energy level, the excess energy will be emitted as a photon,
whose energy corresponds to the difference between the initial and final energy
levels (hf = E2 − E1).
• Photons emitted from such transitions, when allowed to pass through a narrow slit
and a diffraction grating, will give rise to a set of line spectrum, unique to the
element.
High
Potential
Difference
Diffraction
Grating
Line spectrum
Low Pressure Gas
Eg.13:
(Ans: C)
The minimum energy to ionise an atom is the energy required to
A
B
C
D
E
add one electron to the atom.
excite the atom from its ground state to its first excited state.
remove one outermost electron from the atom.
remove one innermost electron from the atom.
remove all the electrons from the atom.
Remark:
This is the ionisation energy of the atom.
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(l)
11.Quantum Physics
Distinguish between emission and absorption line spectra
Emission Line Spectra for Hydrogen
•
•
A spectrum is produced by using a prism or diffraction grating to separate the
various wavelengths in a beam of light. It is the set of wavelengths that are
observed.
A line spectrum contains a discrete set of wavelengths, like the ones shown
here. It is a characteristic feature of an element. Examples of emission line
spectra are as seen above.
Emission
E2
hf =
hc
= E2 − E1
λ
E1
An excited electron transits to a lower energy
level by emitting a photon.
•
Emission refers to the emission of a specific amount of energy from an isolated
atom (e.g. an atom of an element in vapour or gaseous state, at low pressure)
when its electrons jump from higher to lower allowed orbits, releasing the specific
excess energy.
•
Such specific energy E released can only take the form of a photon, with the
corresponding frequency f related by
E = hf =
hc
= E 2 − E1
λ
•
An emission line spectrum of an element consists of a series of separate bright
lines of definite frequencies (or wavelengths) on a dark background. It is
produced when a stream of photons of different frequencies is passed through a
narrow slit and normally through a diffraction grating.
•
These photons are emitted randomly from transitions (from higher to lower
excited states or the ground state) in the excited atoms of the element in a
vapour or gas at low pressure.
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Eg.14:
11.Quantum Physics
•
Since each element has a unique set of orbital electrons, the emission line
spectrum of an element is also unique, enabling it to be used as a means of
identification of the element.
•
Atoms in solid state (e.g. a tungsten filament) are very close to one another, no
longer isolated. The energy levels of each atom are no longer identical to one
another. Emission from these atoms will result in a continuous spectrum (i.e. the
numerous lines diffracted on the screen are so close to each other that
separation between these lines are negligible and the spectrum appears to be
continuous).
(Ans: B)
Which of the following provides experimental evidence for discrete electron energy
levels in atoms?
A
B
C
D
E
the spectrum of a tungsten filament lamp.
the spectrum of a sodium discharge lamp.
the photoelectric effect.
the emission of β-particles by radioactive atoms.
the emission of γ-rays by radioactive atoms.
Remark:
Sodium in a discharge lamp is in gaseous state. Tungsten filament is a solid.
Eg.15:
(Ans: A)
The existence of energy levels within atoms can be demonstrated directly by
observing that
A
B
C
D
E
atoms can emit spectra.
photoelectrons are only emitted for wavelength greater than a critical
wavelength.
some α-particles are reflected back through very large angles by atoms in a
solid.
X-rays with frequencies up to a certain maximum are emitted by a target.
atoms in a solid diffract electrons in the same way as crystals diffract X-rays.
Absorption
•
•
•
Absorption refers to the absorption of energy by an atom from external sources.
Absorption can occur only if it results in the atom being excited to a higher
allowed energy state, i.e. only a specific amount of energy ΔE can be absorbed.
No absorption will occur if too little or too much energy is given to the atom.
E2
ΔE = E 2 − E1
E1
An electron transits to a higher energy
level after absorption.
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11.Quantum Physics
There are 3 ways in which an atom can absorb energy:
(1)
(2)
(3)
from thermal energy (e.g. when an element is heated by a bunsen flame); or
from kinetic energy of a colliding particle
(e.g. when an atom in a vapour in a tube at low pressure is collided by an
electron moving across the tube, say, in an electric fluorescent lamp.); or
from an incident photon (e.g. when a beam of bright light is incident on a
vapour or gas.)
Note:
For (1) and (2), an atom will absorb the exact amount of energy required for a
particular transition, which may be any fraction of the energy supplied by the source.
For (3), only photons with the exact amount of energy required for the transition will
be absorbed. Photons with energies lower or higher than this exact amount will not
be absorbed.
In the diagram below, the bright light beam originates from the filament lamp. The
sodium lamp does not give out light on its own, consisting of only sodium vapour.
The resulting light beams would pass through the collimator which makes the beams
parallel and incident normally onto the diffraction grating. The spectrum formed is
observed using the telescope.
•
An absorption line spectrum of an element consists of a series of separate dark
lines of definite frequencies (or wavelengths) on a coloured background.
•
The coloured background is produced when a stream of photons of different
frequencies from a white light source (e.g. tungsten filament lamp) is passed
through a narrow slit and a diffraction grating.
•
The cool vapour of the element concerned is placed between the white light
source and the narrow slit, such that its atoms may absorb excitation energy from
photons incident from the white light source.
•
The unabsorbed photons from the white light source will be incident on the
screen with the original intensity.
•
After the absorption, the excited atoms will eventually go back to the ground state
by emitting the same photons absorbed earlier.
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11.Quantum Physics
•
The re-emitted photons from the excited atoms in the cool vapour of the element
will be in all directions. Only a small fraction of the re-emitted photons will be
incident onto the screen, hence they will form lines with a much lower intensity.
•
This accounts for the dark lines in the spectrum, which appear on the exact
positions as the corresponding emission line spectrum of the element.
•
Note that these dark lines are not totally dark. In fact, they are of the same colour
as their neighbouring background. They only appear dark against the coloured
background because they are too dim, by comparison.
Summary:
Distinguishing
between
Emission Line Spectra
Absorption Line Spectra
In terms
of
Initial state of
gas atoms
•
produced by hot (gas atoms are
initially at excited state) vapour
or gas.
•
Spectrum
Pattern
•
a series of separate bright lines
of definite wavelength on a dark
background
•
Explanation
•
Excited atoms emit photons
with the characteristic
wavelengths corresponding to
the transitions from higher to
lower energy levels.
•
Cool gas absorbs photons with
the characteristic wavelengths
corresponding to the transitions
from lower to higher energy
levels.
•
Wavelengths corresponding to
bright lines on the spectrum are
characteristics of the element
emitting the light.
•
Subsequently, the excited atoms
transit back to the lower energy
levels by emitting the same
photons in all directions Æ
Small fraction of emitted
photons incident on screen Æ
lines of lower intensity formed
(appear as dark lines on the
spectrum)
•
Wavelengths corresponding to
dark lines on the spectrum are
characteristics of the element
emitting the light.
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produced when white light
passes through cool (gas atoms
are initially at ground state)
vapour or gas.
a series of separate dark lines
of definite wavelength on a
coloured background
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Eg.16:
(Ans: E)
When a parallel beam of white light passes through a metal vapour, dark lines
appear in the spectrum of the emergent light. This is principally because energy is
absorbed and
A
B
C
D
E
Eg.17:
11.Quantum Physics
is not re-emitted at all.
is re-radiated as infra-red.
is re-radiated as ultra-violet.
is re-radiated gradually over a long period of time.
is re-radiated uniformly in all directions.
(Ans: C)
White light from a tungsten filament lamp is passed through sodium vapour and
viewed through a diffraction grating.
Which of the following best describes the spectrum which would be seen?
A
B
C
D
coloured lines on a black background
coloured lines on a white background
dark lines on a coloured background
dark lines on a white background
Remark:
The coloured background is due to the white light from tungsten lamp forming a
continuous spectrum.
(m)
Recall and solve problems using the relation hf = E1 − E2.
In a transition from an initial level of energy E1 to a final level of energy E2 ( E1 > E2 ),
a photon of frequency f is emitted, with its energy E given by
E = hf = h
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c
= E1 − E2
λ
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Eg.18:
11.Quantum Physics
(Ans: D)
An atom emits a spectral line of wavelength λ when an electron makes a transition
between the levels E1 and E2. Which expression correctly relates λ, E1 and E2?
h
A λ = ( E1 − E2)
B λ = ch ( E1 − E2)
c
c
ch
C λ=
D λ=
h(E1 − E 2 )
E1 − E 2
Answer:
hf = E1 − E2
Eg.19:
Î
h
c
= E1 − E2 Î
λ
λ=
hc
E1 − E 2
(Ans: A)
In the figure below, E1 to E6 represent some of the energy levels of an electron in the
hydrogen atom.
E6
E5
E4
−0.38 eV
−0.54 eV
−0.85 eV
E3
−1.5 eV
E2
−3.4 eV
E1
−13.6 eV
Which one of the following transitions produces a photon of wavelength in the ultraviolet region of the electro-magnetic spectrum? [ 1 eV = 1.6×10-19 J. ]
A E2 Æ E1
D E5 Æ E4
B E3 Æ E2
E E6 Æ E5
C E4 Æ E3
Answer:
Assume a transition from E2 to E1, the resulting photon emitted would have an
energy, E = (−3.4 eV) − (−13.6 eV) = 10.2 eV = 10.2(1.6×10-19 J) = 1.63×10-18 J.
The frequency and wavelength of the photon are related by the equation
c
E = hf = h
λ
3 × 10 8 ms -1
c
Î
λ = h = (6.63×10-34 Js)
E
1.63 × 10 −18 J
= 1.22×10-7 m
This corresponds to the wavelength of ultra-violet radiation.
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8866 H1 Physics – J2/2011
Eg.20:
11.Quantum Physics
(Ans: E)
The diagram below represents, drawn to scale, the energy levels for an electron in a
certain atom.
E4
E3
E2
energy
E1
The transition from E3 to E1 produces a green line. What transition could give rise to
a red line?
A E4 to E3
D E3 to E2
B E4 to E2
E E2 to E1
C E4 to E1
Remark:
• Frequency of red light is lower than frequency of green light, so energy of a
photon from a red light is lower than that from a green light.
•
But gaps between energy levels are bigger for lower levels than for higher levels.
•
The transition from E3 to E1 gives a green light (in the visible light region of the
em spectrum), then to give a red light, the transition must also be to E1.
•
Transition to other levels would give a photon of much lower energy, regardless
of where the transition is originated from.
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8866 H1 Physics – J2/2011
11.Quantum Physics
Annex A: Results and explanations for the photoelectric effect
Result
1
Experimental
Observation
The rate of emission of
photoelectrons
( measured as
photoelectric current ) is
proportional to the incident
radiation intensity.
Intensity is the rate of
incidence of energy per unit
area.
Explanation
Considering light as a wave
Considering light as a particle
A beam of light carries a continuous flow of energy
when it is incident onto the metal surface. The
intensity of the beam is proportional to the square of
the wave amplitude.
Since each photon has a specific amount of energy (= hf),
then the intensity of a beam of light is a measure of the rate
of incidence of photons, i.e. the number of photons incident
per unit time.
Intensity is a measure of the rate at which energy
would be imparted onto the electrons per unit
surface area of the emitting electrode C.
The higher the intensity, the more photons per unit time is
incident onto the emitting electrode C.
The higher the intensity of the light beam, the higher
the energy imparted onto the electrons. This
resulted in a higher number of photoelectrons
emitted per unit time which means a higher current.
photoelectric
current
It is assumed each photon gives all its energy (= hf) to a
single electron in the metal.
Hence the more photons incident onto the metal surface,
the more photoelectrons emitted.
Can explain observation.
Can explain observation.
Intensity
2
For every material of C
irradiated, there is a
minimum frequency (
threshold frequency fo )
required to liberate an
electron from the surface.
Emission of photoelectrons from the metal surface
should occur at any frequency. It is independent of
frequency.
Cannot explain observation.
This fo is independent of the
light intensity.
There is a certain minimum energy called the work
function energy, φ necessary to liberate an electron from
its surface. Since the energy of a photon is expressed as
hf, hence the minimum photon energy required to liberate
an electron from its surface must be hfo (= φ) where fo is the
threshold frequency.
fo andφ are characteristic of the material C.
Can explain observation.
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8866 H1 Physics – J2/2011
Explanation
Experimental
Observation
Result
3
11.Quantum Physics
The maximum kinetic
( photoelectric ) energy of
emitted electrons depends
only on the frequency of the
incident radiation, and not
its intensity.
current
Frequency fixed,
Intensity I2 > I1
Considering light as a wave
The higher the intensity, the higher the energy
imparted onto the electrons on the metal surface.
Considering light as a particle
The emitted photoelectrons would have absorbed energy
from the incident photons.
The electrons should then be emitted with higher
kinetic energies, which contradict the experimental
result.
Since each electron absorbs energy only from one photon,
what matters is the energy of each photon (which depends
only on its frequency).
Cannot explain observation.
High intensity means there is a large number of photons
incident onto the electrode C per unit time, but this does not
affect the energy absorbed by a single electron.
I2
I1
Can explain observation.
V
Vs
current
Intensity fixed,
Frequency f2 > f1
f2
f1
VS2 VS1
4
V
The emission of
photoelectrons starts with
no observable time lag,
even for very low intensity
of incident radiation.
For very low intensity radiation, an electron would
require some time to absorb enough energy to be
emitted from the metal surface. Then there would
be a certain time lag in its emission.
Cannot explain observation.
For the incident photon, its energy will be instantaneously
transferred to the absorbing electron, in a one-to-one
interaction.
This accounts for the instantaneous emission of the
photoelectron.
Can explain observation.
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