Chapter 11 Dual Nature of Radiation and Matter
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Transcript of Chapter 11 Dual Nature of Radiation and Matter
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Question 11.1:
Find the
maximum frequency, and
minimum wavelength of X-rays produced by 30 kV electrons.
Answer
Potential of the electrons, V = 30 kV = 3 × 104 V
Hence, energy of the electrons, E = 3 × 104 eV
Where,
e = Charge on an electron = 1.6 × 10−19 C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E = hν
Where,
h = Planck’s constant = 6.626 × 10−34 Js
Hence, the maximum frequency of X-rays produced is
(b)The minimum wavelength produced by the X-rays is given as:
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Hence, the minimum wavele
Question 11.2:
The work function of caesiuincident on the metal surface
maximum kinetic energy of t
Stopping potential, and
maximum speed of the emitt
Answer
Work function of caesium m
Frequency of light,
(a)The maximum kinetic ene
Where,
h = Planck’s constant = 6.62
gth of X-rays produced is 0.0414 nm.
metal is 2.14 eV. When light of frequency 6 ×, photoemission of electrons occurs. What is the
he emitted electrons,
d photoelectrons?
etal,
rgy is given by the photoelectric effect as:
× 10−34 Js
1014 Hz is
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Photoelectric cut-off voltage,
The maximum kinetic energ
Where,
e = Charge on an electron =
Therefore, the maximum kinexperiment is 2.4 × 10−19 J.
Question 11.4:
Monochromatic light of wav power emitted is 9.42 mW.
Find the energy and moment
How many photons per seco(Assume the beam to have u
How fast does a hydrogen atthat of the photon?
Answer
Wavelength of the monochr
Power emitted by the laser,
V 0 = 1.5 V
of the emitted photoelectrons is given as:
.6 × 10−19 C
etic energy of the photoelectrons emitted in the
length 632.8 nm is produced by a helium-neon
m of each photon in the light beam,
d, on the average, arrive at a target irradiated biform cross-section which is less than the targe
m have to travel in order to have the same mo
matic light, λ = 632.8 nm = 632.8 × 10−9 m
= 9.42 mW = 9.42 × 10−3 W
given
laser. The
this beam?area), and
entum as
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Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg
(a)The energy of each photon is given as:
The momentum of each photon is given as:
(b) Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where,
v = Speed of the hydrogen atom
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Therefore, every second,
Question 11.6:
In an experiment on photoelof incident light is found to b
Answer
The slope of the cut-off volta
Where,
e = Charge on an electron =
h = Planck’s constant
Therefore, the value of Planc
Question 11.7:
A 100 W sodium lamp radiatthe centre of a large sphere twavelength of the sodium ligwith the sodium light? (b) At
photons are incident per square metre o
ctric effect, the slope of the cut-off voltage verse 4.12 × 10−15 V s. Calculate the value of Planc
ge (V ) versus frequency (ν) of an incident light
.6 × 10−19 C
k’s constant is
es energy uniformly in all directions. The lampat absorbs all the sodium light which is incidenht is 589 nm. (a) What is the energy per photonwhat rate are the photons delivered to the sphe
n earth.
us frequency’s constant.
is given as:
is located att on it. Theassociatede?
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Answer
Power of the sodium lamp, P
Wavelength of the emitted s
Planck’s constant, h = 6.626
Speed of light, c = 3 × 108 m
(a)The energy per photon as
(b) Number of photons delive
The equation for power can
Therefore, every second,
Question 11.8:
= 100 W
dium light, λ = 589 nm = 589 × 10−9 m
× 10−34 Js
s
ociated with the sodium light is given as:
red to the sphere = n
e written as:
photons are delivered to the sphere.
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The threshold frequency for1014 Hz is incident on the me
Answer
Threshold frequency of the
Frequency of light incident o
Charge on an electron, e = 1.
Planck’s constant, h = 6.626
Cut-off voltage for the photo
The equation for the cut-off
Therefore, the cut-off voltag
Question 11.9:
The work function for a certaemission for incident radiati
Answer
certain metal is 3.3 × 1014 Hz. If light of frequ tal, predict the cutoff voltage for the photoelect
etal,
n the metal,
6 × 10−19 C
× 10−34 Js
electric emission from the metal =
nergy is given as:
for the photoelectric emission is
in metal is 4.2 eV. Will this metal give photoeln of wavelength 330 nm?
ncy 8.2 ×ic emission.
ctric
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No
Work function of the metal,
Charge on an electron, e = 1.
Planck’s constant, h = 6.626
Wavelength of the incident r
Speed of light, c = 3 × 108 m
The energy of the incident p
It can be observed that the eof the metal. Hence, no phot
Question 11.10:
Light of frequency 7.21 × 10maximum speed of 6.0 × 105
frequency for photoemission
Answer
Frequency of the incident ph
Maximum speed of the elect
6 × 10−19 C
× 10−34 Js
diation, λ = 330 nm = 330 × 10−9 m
s
oton is given as:
ergy of the incident radiation is less than the welectric emission will take place.
14 Hz is incident on a metal surface. Electronsm/s are ejected from the surface. What is the thof electrons?
oton,
ons, v = 6.0 × 105 m/s
rk function
ith areshold
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Planck’s constant, h = 6.626
Mass of an electron, m = 9.1
For threshold frequency ν0, t
Therefore, the threshold freq
Question 11.11:
Light of wavelength 488 nm photoelectric effect. When listopping (cut-off) potential omaterial from which the emi
Answer
Wavelength of light produce
= 488 × 10−9 m
Stopping potential of the pho
1eV = 1.6 × 10−19 J
∴ V 0 =
Planck’s constant, h = 6.6 ×
× 10−34 Js
× 10−31 kg
e relation for kinetic energy is written as:
ency for the photoemission of electrons is
is produced by an argon laser which is used in tht from this spectral line is incident on the emi
f photoelectrons is 0.38 V. Find the work functiter is made.
by the argon laser, λ = 488 nm
toelectrons, V 0 = 0.38 V
0−34 Js
heter, theon of the
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Charge on an electron, e = 1.
Speed of light, c = 3 × 10 m/
From Einstein’s photoelectriof the material of the emitter
Therefore, the material with
Question 11.12:
Calculate the
momentum, and
de Broglie wavelength of the
Answer
Potential difference, V = 56
Planck’s constant, h = 6.6 ×
Mass of an electron, m = 9.1
Charge on an electron, e = 1.
6 × 10−19 C
effect, we have the relation involving the woras:
hich the emitter is made has the work function
electrons accelerated through a potential differ
0−34 Js
× 10−31 kg
6 × 10−19 C
function Φ0
of 2.16 eV.
nce of 56 V.
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At equilibrium, the kinetic ei.e., we can write the relation
The momentum of each acce
p = mv
= 9.1 × 10−31 × 4.44 × 106
= 4.04 × 10−24 kg m s−1
Therefore, the momentum of
De Broglie wavelength of anrelation:
Therefore, the de Broglie wa
Question 11.13:
What is the
momentum,
speed, and
ergy of each electron is equal to the acceleratinfor velocity (v) of each electron as:
lerated electron is given as:
each electron is 4.04 × 10−24 kg m s−1.
electron accelerating through a potential V , is g
elength of each electron is 0.1639 nm.
g potential,
iven by the
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de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer
Kinetic energy of the electron, E k = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
For the electron, we can write the relation for kinetic energy as:
Where,
v = Speed of the electron
Momentum of the electron, p = mv
= 9.1 × 10−31 × 6.496 × 106
= 5.91 × 10−24 kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1.
Speed of the electron, v = 6.496 × 106 m/s
De Broglie wavelength of an electron having a momentum p, is given as:
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Therefore, the de Broglie wa
Question 11.14:
The wavelength of light frokinetic energy at which
an electron, and
a neutron, would have the sa
Answer
Wavelength of light of a sodi
Mass of an electron, me= 9.1
Mass of a neutron, mn= 1.66
Planck’s constant, h = 6.6 ×
For the kinetic energy K , of arelation:
We have the relation for de
elength of the electron is 0.112 nm.
the spectral emission line of sodium is 589 nm
e de Broglie wavelength.
um line, λ = 589 nm = 589 × 10−9 m
× 10−31 kg
× 10−27 kg
0−34 Js
n electron accelerating with a velocity v, we ha
roglie wavelength as:
. Find the
e the
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Substituting equation (2) in e
Hence, the kinetic energy of
Using equation (3), we can
Hence, the kinetic energy of
Question 11.15:
What is the de Broglie wavel
quation (1), we get the relation:
he electron is 6.9 × 10−25J or 4.31 μeV.
rite the relation for the kinetic energy of the ne
he neutron is 3.78 × 10−28 J or 2.36 neV.
ength of
tron as:
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a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
Answer
(a)Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 × 10−34 Js
De Broglie wavelength of the bullet is given by the relation:
Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
(c)Mass of the dust particle, m = 1 × 10−9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
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Question 11.16:An electron and a photon eac
their momenta,
the energy of the photon, an
the kinetic energy of electro
Answer
Wavelength of an electron
= 1 × 10−9 m
Planck’s constant, h = 6.63 ×
The momentum of an eleme
It is clear that momentum dewavelengths of an electron a
The energy of a photon is gi
h have a wavelength of 1.00 nm. Find
.
10−34 Js
tary particle is given by de Broglie relation:
ends only on the wavelength of the particle. Sid a photon are equal, both have an equal mom
en by the relation:
ce thentum.
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Where,
Speed of light, c = 3 × 108 m
Therefore, the energy of the
The kinetic energy ( K ) of an
Where,
m = Mass of the electron = 9.
p = 6.63 × 10−25 kg m s−1
Hence, the kinetic energy of
Question 11.17:
For what kinetic energy of a10−10 m?
Also find the de Broglie wav
having an average kinetic en
Answer
s
hoton is 1.243 keV.
electron having momentum p,is given by the rel
.1 × 10−31 kg
he electron is 1.51 eV.
neutron will the associated de Broglie wavelen
elength of a neutron, in thermal equilibrium wit
rgy of (3/2) kT at 300 K.
ation:
th be 1.40 ×
h matter,
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De Broglie wavelength of the neutron, λ = 1.40 × 10−10 m
Mass of a neutron, mn = 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34
Js
Kinetic energy ( K ) and velocity (v) are related as:
… (1)
De Broglie wavelength ( λ) and velocity (v) are related as:
Using equation (2) in equation (1), we get:
Hence, the kinetic energy of the neutron is 6.75 × 10−21 J or 4.219 × 10−2 eV.
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 × 10−23 kg m2 s−2 K −1
Average kinetic energy of the neutron:
The relation for the de Broglie wavelength is given as:
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Therefore, the de Broglie wa
Question 11.18:
Show that the wavelength ofwavelength of its quantum (p
Answer
The momentum of a photon
Where,
λ = Wavelength of the electr
c = Speed of light
h = Planck’s constant
elength of the neutron is 0.146 nm.
electromagnetic radiation is equal to the de Brohoton).
aving energy (hν) is given as:
magnetic radiation
glie
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De Broglie wavelength of th
Where,
m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred froelectromagnetic radiation is
Question 11.19:
What is the de Broglie wavelthe molecule is moving with(Atomic mass of nitrogen =
Answer
Temperature of the nitrogen
Atomic mass of nitrogen = 1
Hence, mass of the nitrogen
But 1 u = 1.66 × 10−27 kg
∴m = 28.0152 ×1.66 × 10−27
Planck’s constant, h = 6.63 ×
Boltzmann constant, k = 1.3
photon is given as:
equations (i) and (ii) that the wavelength of thqual to the de Broglie wavelength of the photo
ength of a nitrogen molecule in air at 300 K? Athe root-mean square speed of molecules at this4.0076 u)
olecule, T = 300 K
.0076 u
olecule, m = 2 × 14.0076 = 28.0152 u
g
10−34 Js
× 10−23 J K −1
e.
ssume thattemperature.
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We have the expression that
with the root mean square sp
Hence, the de Broglie wavel
Therefore, the de Broglie wa
Question 11.20:
Estimate the speed with whitube impinge on the collectothe emitter. Ignore the smallelectron, i.e., its e/m is given
Use the same formula you eof 10 MV. Do you see what i
Answer
(a)Potential difference acros
elates mean kinetic energy of the nitro
eed as:
ngth of the nitrogen molecule is given as:
elength of the nitrogen molecule is 0.028 nm.
h electrons emitted from a heated emitter of anmaintained at a potential difference of 500 V
initial speeds of the electrons. The specific char
to be 1.76 × 1011 C kg−1.
ploy in (a) to obtain electron speed for an colles wrong? In what way is the formula to be mod
the evacuated tube, V = 500 V
en molecule
evacuatedith respect toe of the
ctor potentialfied?
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Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electron is
(b)Potential of the anode, V = 10 MV = 10 × 106
V
The speed of each electron is given as:
This result is wrong because nothing can move faster than light. In the above formula, theexpression (mv
2/2) for energy can only be used in the non-relativistic limit, i.e., for v <<c.
For very high speed problems, relativistic equations must be considered for solving them.In the relativistic limit, the total energy is given as:
E = mc2
Where,
m = Relativistic mass
m0 = Mass of the particle at rest
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Kinetic energy is given as:
K = mc2
− m0c2
Question 11.21:
A monoenergetic electron bemagnetic field of 1.30 × 10−4
circle traced by the beam, gi
Is the formula you employ inelectron beam? If not, in wha
[Note: Exercises 11.20(b) an beyond the scope of this boo point that the formulas you u
speeds or energies. See answmeans.]
Answer
(a)Speed of an electron, v =
Magnetic field experienced b
Specific charge of an electro
Where,
e = Charge on the electron =
m = Mass of the electron = 9.
The force exerted on the elec
θ = Angle between the magn
The magnetic field is normal
am with electron speed of 5.20 × 106 m s−1 is suT normal to the beam velocity. What is the raden e/m for electron equals 1.76 × 1011 C kg−1.
(a) valid for calculating radius of the path of at way is it modified?
d 11.21(b) take you to relativistic mechanics w. They have been inserted here simply to emph
se in part (a) of the exercises are not valid at ve
ers at the end to know what ‘very high speed or
.20 × 106 m/s
y the electron, B = 1.30 × 10−4 T
, e/m = 1.76 × 1011 C kg−1
1.6 × 10−19 C
.1 × 10−31 kg−1
tron is given as:
etic field and the beam velocity
to the direction of beam.
bject to aus of the
0 MeV
ich isasise they high
energy’
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The beam traces a circular path of radius, r . It is the magnetic field, due to its bending
nature, that provides the centripetal force for the beam.
Hence, equation (1) reduces to:
Therefore, the radius of the circular path is 22.7 cm.
Energy of the electron beam, E = 20 MeV
The energy of the electron is given as:
This result is incorrect because nothing can move faster than light. In the above formula,the expression (mv
2/2) for energy can only be used in the non-relativistic limit, i.e., for v
<< c
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
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Where,
= Mass of the particle at
Hence, the radius of the circ
Question 11.22:
An electron gun with its coll bulb containing hydrogen ga2.83 × 10−4 T curves the path
path can be viewed because telectrons, and emitting lighttube’ method. Determine e/m
Answer
Potential of an anode, V = 10
Magnetic field experienced b
Radius of the circular orbit r
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron i
est
lar path is given as:
ctor at a potential of 100 V fires out electrons iat low pressure (∼10−2 mm of Hg). A magnetiof the electrons in a circular orbit of radius 12.
he gas ions in the path focus the beam by attracy electron capture; this method is known as thfrom the data.
0 V
y the electrons, B = 2.83 × 10−4 T
= 12.0 cm = 12.0 × 10−2
m
s equal to its kinetic energy, i.e.,
a sphericalfield ofcm. (The
ing‘fine beam
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It is the magnetic field, due t
for the beam. H
Centripetal force = Magnetic
Putting the value of v in equa
Therefore, the specific charg
Question 11.23:
An X-ray tube produces a coat 0.45 Å. What is the maxi
From your answer to (a), guerequired in such a tube?
Answer
its bending nature, that provides the centripeta
nce, we can write:
force
tion (1), we get:
e ratio (e/m) is
tinuous spectrum of radiation with its short waum energy of a photon in the radiation?
ss what order of accelerating voltage (for electr
l force
velength end
ons) is
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Wavelength produced by an
Planck’s constant, h = 6.626
Speed of light, c = 3 × 108 m
The maximum energy of a p
Therefore, the maximum ene
Accelerating voltage provideray of 27.6 keV, the incidentenergy. Hence, an acceleratiX-rays.
Question 11.24:
In an accelerator experimentcertain event is interpreted aBeV into two γ-rays of equal
(1BeV = 109 eV)
Answer
Total energy of two γ-rays:
E = 10. 2 BeV
= 10.2 × 109 eV
= 10.2 × 109 × 1.6 × 10−10 J
-ray tube,
× 10−34 Js
s
oton is given as:
rgy of an X-ray photon is 27.6 keV.
s energy to the electrons for producing X-rays.electrons must possess at least 27.6 keV of king voltage of the order of 30 keV is required for
on high-energy collisions of electrons with posiannihilation of an electron-positron pair of totaenergy. What is the wavelength associated wit
o get an X- tic electric
producing
trons, al energy 10.2each γ-ray?
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Hence, the energy of each γ-
Planck’s constant,
Speed of light,
Energy is related to wavelen
Therefore, the wavelength as
Question 11.25:
Estimating the following twyou why radio engineers dotells you why our eye can ne
The number of photons emit power, emitting radiowaves
The number of photons enterminimum intensity of white larea of the pupil to be about6 × 1014 Hz.
Answer
Power of the medium wave t
ay:
th as:
sociated with each γ-ray is
numbers should be interesting. The first numbot need to worry much about photons! The secer ‘count photons’, even in barely detectable li
ed per second by a Medium wave transmitter of wavelength 500 m.
ing the pupil of our eye per second correspondiight that we humans can perceive (∼10−10 W m
.4 cm2, and the average frequency of white lig
ansmitter, P = 10 kW = 104 W = 104 J/s
r will tellnd numberht.
10 kW
g to the2). Take thet to be about
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Hence, energy emitted by the transmitter per second, E = 104
Wavelength of the radio wave, λ = 500 m
The energy of the wave is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
Let n be the number of photons emitted by the transmitter.
∴nE 1 = E
The energy ( E 1) of a radio photon is very less, but the number of photons (n) emitted persecond in a radio wave is very large.
The existence of a minimum quantum of energy can be ignored and the total energy of aradio wave can be treated as being continuous.
Intensity of light perceived by the human eye, I = 10−10 W m−2
Area of a pupil, A = 0.4 cm
2
= 0.4 × 10
−4
m
2
Frequency of white light, ν= 6 × 1014 Hz
The energy emitted by a photon is given as:
E = h ν
Where,
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h = Planck’s constant = 6.6 ×
∴ E = 6.6 × 10−34 × 6 × 1014
= 3.96 × 10−19 J
Let n be the total number of
The total energy per unit for
E = n × 3.96 × 10−19 J s−1 m−
The energy per unit area per
∴ E = I
n × 3.96 × 10−19 = 10−10
= 2.52 × 108 m2 s−1
The total number of photons
n A = n × A
= 2.52 × 108 × 0.4 × 10−4
= 1.008 × 104 s−1
This number is not as large ahuman eye to never see the i
Question 11.26:
Ultraviolet light of wavelengcell made of molybdenum mfunction of the metal. Howred light of wavelength 6328
Answer
10−34 Js
hotons falling per second, per unit area of the p
n falling photons is given as:
second is the intensity of light.
entering the pupil per second is given as:
s the one found in problem (a), but it is large endividual photons.
th 2271 Å from a 100 W mercury source irradiaetal. If the stopping potential is −1.3 V, estimat
ould the photo-cell respond to a high intensityÅ produced by a He-Ne laser?
upil.
ough for the
tes a photo- the work
∼105 W m−2)
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Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m
Stopping potential of the metal, V 0 = 1.3 V
Planck’s constant, h = 6.6 × 10−34
J
Charge on an electron, e = 1.6 × 10−19 C
Work function of the metal =
Frequency of light = ν
We have the photo-energy relation from the photoelectric effect as:
= hν − eV 0
Let ν0 be the threshold frequency of the metal.
∴ = hν0
Wavelength of red light, = 6328 × 10−10 m
∴Frequency of red light,
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Since ν0> νr , the photocell wi
Question 11.27:
Monochromatic radiation ofirradiates photosensitive matmeasured to be 0.54 V. Theirradiates the same photo-cel
Answer
Wavelength of the monochr
= 640.2 × 10−9 m
Stopping potential of the neo
Charge on an electron, e = 1.
Planck’s constant, h = 6.6 ×
Let be the work function
We have the photo-energy re
eV 0 = hν −
Wavelength of the radiation
ll not respond to the red light produced by the l
avelength 640.2 nm (1nm = 10−9
m) from a neerial made of caesium on tungsten. The stoppinource is replaced by an iron source and its 427.l. Predict the new stopping voltage.
matic radiation, λ = 640.2 nm
n lamp, V 0 = 0.54 V
6 × 10−19 C
0−34 Js
nd ν be the frequency of emitted light.
lation from the photoelectric effect as:
mitted from an iron source, λ' = 427.2 nm
ser.
on lampvoltage isnm line
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= 427.2 × 10−9 m
Let be the new stopping p
Hence, the new stopping pot
Question 11.28:
A mercury lamp is a conveni photoelectric emission, sincethe red end of the visible spefollowing lines from a mercu
λ1 = 3650 Å, λ2= 4047 Å, λ3
The stopping voltages, respe
V 01 = 1.28 V, V 02 = 0.95 V,
Determine the value of Plancfor the material.
[ Note: You will notice that tcan take to be 1.6 × 10−19 C).
by Millikan, who, using hisEinstein’s photoelectric equathe value of h.]
Answer
otential. Hence, photo-energy is given as:
ntial is 1.50 eV.
ent source for studying frequency dependence oit gives a number of spectral lines ranging fro
ctrum. In our experiment with rubidium photo-cry source were used:
4358 Å, λ4= 5461 Å, λ5= 6907 Å,
tively, were measured to be:
03 = 0.74 V, V 04 = 0.16 V, V 05 = 0 V
k’s constant h, the threshold frequency and wor
get h from the data, you will need to know e (Experiments of this kind on Na, Li, K, etc. werwn value of e (from the oil-drop experiment) ction and at the same time gave an independent
fthe UV to
ell, the
k function
hich youe performednfirmedstimate of
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Einstein’s photoelectric equation is given as:
eV 0 = hν−
Where,
V 0 = Stopping potential
h = Planck’s constant
e = Charge on an electron
ν = Frequency of radiation
= Work function of a material
It can be concluded from equation (1) that potential V 0 is directly proportional tofrequency ν.
Frequency is also given by the relation:
This relation can be used to obtain the frequencies of the various lines of the givenwavelengths.
The given quantities can be listed in tabular form as:
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Frequency × 1014
Hz 8.219 7.412 6.884 5.493 4.343
Stopping potential V 0 1.28 0.95 0.74 0.16 0
The following figure shows a graph between νand V 0.
It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 ×1014 Hz, which is the threshold frequency (ν0) of the material. Point D corresponds to afrequency less than the threshold frequency. Hence, there is no photoelectric emission forthe λ5 line, and therefore, no stopping voltage is required to stop the current.
Slope of the straight line =
From equation (1), the slope can be written as:
The work function of the metal is given as:
= h ν0
= 6.573 × 10−34 × 5 × 1014
= 3.286 × 10−19 J
= 2.054 eV
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Question 11.29:
The work function for the fol
Na: 2.75 eV; K: 2.30 eV; M photoelectric emission for am away from the photocell?away?
Answer
Mo and Ni will not show ph
Wavelength for a radiation,
Speed of light, c = 3 × 108 m
Planck’s constant, h = 6.6 ×
The energy of incident radiat
It can be observed that the efunction of Na and K only. It
photoelectric emission.
If the source of light is brougthen the intensity of radiatioradiation. Hence, the resultemitted from Na and K will i
Question 11.30:
lowing metals is given:
: 4.17 eV; Ni: 5.15 eV. Which of these metalsadiation of wavelength 3300 Å from a He-Cd l
hat happens if the laser is brought nearer and
toelectric emission in both cases
= 3300 Å = 3300 × 10−10 m
s
0−34 Js
ion is given as:
ergy of the incident radiation is greater than theis less for Mo and Ni. Hence, Mo and Ni will n
ht near the photocells and placed 50 cm away fwill increase. This does not affect the energy o
ill be the same as before. However, the photoelncrease in proportion to intensity.
ill not giveser placed 1laced 50 cm
workot show
om them,f theectrons
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Light of intensity 10−5 W m−2 falls on a sodium photo-cell of surface area 2 cm2.Assuming that the top 5 layers of sodium absorb the incident energy, estimate timerequired for photoelectric emission in the wave-picture of radiation. The work functionfor the metal is given to be about 2 eV. What is the implication of your answer?
Answer
Intensity of incident light, I = 10−5 W m−2
Surface area of a sodium photocell, A = 2 cm2 = 2 × 10−4 m2
Incident power of the light, P = I × A
= 10−5 × 2 × 10−4
= 2 × 10−9 W
Work function of the metal, = 2 eV
= 2 × 1.6 × 10−19
= 3.2 × 10−19 J
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, Ae is 10−20 m2.
Hence, the number of conduction electrons in n layers is given as:
The incident power is uniformly absorbed by all the electrons continuously. Hence, theamount of energy absorbed per second per electron is:
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Time required for photoelect
The time required for the ph practical. Hence, the wave pi
Question 11.31:
Crystal diffraction experimethrough appropriate voltage.comparison, take the waveleatomic spacing in the lattice)
Answer
An X-ray probe has a greate
Wavelength of light emitted
Mass of an electron, me = 9.1
Planck’s constant, h = 6.6 ×
Charge on an electron, e = 1.
The kinetic energy of the ele
Where,
v = Velocity of the electron
ic emission:
toelectric emission is nearly half a year, whichcture is in disagreement with the given experim
ts can be performed using X-rays, or electronsWhich probe has greater energy? (For quantitatgth of the probe equal to 1 Å, which is of the o(me= 9.11 × 10−31 kg).
energy than an electron probe for the same wa
from the probe, λ = 1 Å = 10−10 m
1 × 10−31 kg
0−34 Js
6 × 10−19 C
tron is given as:
is notent.
acceleratedveder of inter-
elength.
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mev = Momentum ( p) of the
According to the de Broglie
Energy of a photon,
Hence, a photon has a greate
Question 11.32:
Obtain the de Broglie wavelseen in Exercise 11.31, an elexperiments. Would a neutro(mn= 1.675 × 10−27 kg)
Obtain the de Broglie wavel(27 ºC). Hence explain whyenvironment before it can be
Answer
De Broglie wavelength =experiment
lectron
rinciple, the de Broglie wavelength is given as:
energy than an electron for the same waveleng
ngth of a neutron of kinetic energy 150 eV. Asctron beam of this energy is suitable for crystal
n beam of the same energy be equally suitable?
ngth associated with thermal neutrons at roomfast neutron beam needs to be thermalised wit
used for neutron diffraction experiments.
; neutron is not suitable for the dif
th.
you havediffractionExplain.
emperaturethe
fraction
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Kinetic energy of the neutron, K = 150 eV
= 150 × 1.6 × 10−19
= 2.4 × 10−17 J
Mass of a neutron, mn = 1.675 × 10−27
kg
The kinetic energy of the neutron is given by the relation:
Where,
v = Velocity of the neutron
mnv = Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å,i.e., 10−10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, aneutron beam of energy150 eV is not suitable for diffraction experiments.
De Broglie wavelength =
Room temperature, T = 27°C = 27 + 273 = 300 K
The average kinetic energy of the neutron is given as:
Where,
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k = Boltzmann constant = 1.
The wavelength of the neutr
This wavelength is comparabenergy neutron beam should
Question 11.33:
An electron microscope usesde Broglie wavelength assocaperture, etc.) are taken to beelectron microscope compar
Answer
Electrons are accelerated by
Charge on an electron, e = 1.
Mass of an electron, me = 9.1
Wavelength of yellow light
The kinetic energy of the ele
E = eV
= 1.6 × 10−19 × 50 × 103
= 8 × 10−15 J
De Broglie wavelength is giv
8 × 10−23 J mol−1 K −1
n is given as:
le to the inter-atomic spacing of a crystal. Hencfirst be thermalised, before using it for diffracti
electrons accelerated by a voltage of 50 kV. Deated with the electrons. If other factors (such asroughly the same, how does the resolving powwith that of an optical microscope which uses
a voltage, V = 50 kV = 50 × 103 V
6 × 10−19 C
1 × 10−31 kg
5.9 × 10−7 m
tron is given as:
en by the relation:
e, the high- n.
termine thenumerical
r of anellow light?
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This wavelength is nearly 10
The resolving power of a miused. Thus, the resolving pooptical microscope.
Question 11.34:
The wavelength of a probe isin some detail. The quark strscale of 10−15 m or less. Thiselectron beams produced byhave been the order of energ0.511 MeV.)
Answer
Wavelength of a proton or a
Rest mass energy of an elect
m0c2 = 0.511 MeV
= 0.511 × 106 × 1.6 × 10−19
= 0.8176 × 10−13
J
Planck’s constant, h = 6.6 ×
Speed of light, c = 3 × 108 m
The momentum of a proton
times less than the wavelength of yellow light.
roscope is inversely proportional to the waveleer of an electron microscope is nearly 105 time
roughly a measure of the size of a structure thacture of protons and neutrons appears at the mistructure was first probed in early 1970’s using
a linear accelerator at Stanford, USA. Guess whof these electron beams. (Rest mass energy of
eutron, λ ≈ 10−15 m
on:
0−34 Js
s
r a neutron is given as:
gth of lights that of an
t it can probenute length-
high energyat mightelectron =
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The relativistic relation for e
Thus, the electron energy emorder of 1.24 BeV.
Question 11.35:
Find the typical de Broglie wtemperature (27 ºC) and 1 attwo atoms under these condi
Answer
De Broglie wavelength assoc
Room temperature, T = 27°C
Atmospheric pressure, P = 1
ergy ( E ) is given as:
itted from the accelerator at Stanford, USA mig
avelength associated with a He atom in helium pressure; and compare it with the mean separ
ions.
iated with He atom =
= 27 + 273 = 300 K
atm = 1.01 × 105 Pa
ht be of the
gas at roomtion between
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Atomic weight of a He atom = 4
Avogadro’s number, NA = 6.023 × 1023
Boltzmann constant, k = 1.38 × 10−23 J mol−1 K −1
Average energy of a gas at temperature T ,is given as:
De Broglie wavelength is given by the relation:
Where,
m = Mass of a He atom
We have the ideal gas formula:
PV = RT
PV = kNT
Where,
V = Volume of the gas
N = Number of moles of the gas
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Mean separation between tw
Hence, the mean separationwavelength.
Question 11.36:
Compute the typical de Brogcompare it with the mean se be about 2 × 10−10 m.
[ Note: Exercises 11.35 and 1gaseous molecules under ord
packets in a metal strongly omolecules in an ordinary gasdistinguished apart from oneimplications which you will
Answer
Temperature, T = 27°C = 27
Mean separation between tw
De Broglie wavelength of an
Where,
h = Planck’s constant = 6.6 ×
atoms of the gas is given by the relation:
etween the atoms is much greater than the de B
lie wavelength of an electron in a metal at 27 ºCaration between two electrons in a metal which
1.36 reveal that while the wave-packets associainary conditions are non-overlapping, the electrerlap with one another. This suggests that whecan be distinguished apart, electrons in a metalanother. This indistinguishibility has many fun
xplore in more advanced Physics courses.]
+ 273 = 300 K
electrons, r = 2 × 10−10 m
electron is given as:
10−34 Js
roglie
andis given to
ed withon wave-
eascannot beamental
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m = Mass of an electron = 9.
k = Boltzmann constant = 1.
Hence, the de Broglie wavelseparation.
Question 11.37:
Answer the following questi
Quarks inside protons and ne(−1/3)e]. Why do they not sh
What is so special about theseparately?
Why should gases be insulat pressures?
Every metal has a definite wthe same energy if incident r
distribution of photoelectron
The energy and momentumthe associated matter wave b
E = hν, p =
But while the value of λ is pof the phase speed νλ) has no
Answer
11 × 10−31 kg
8 × 10−23 J mol−1 K −1
ngth is much greater than the given inter-electr
ns:
utrons are thought to carry fractional charges [( ow up in Millikan’s oil-drop experiment?
ombination e/m? Why do we not simply talk o
rs at ordinary pressures and start conducting at
rk function. Why do all photoelectrons not codiation is monochromatic? Why is there an ene
?
f an electron are related to the frequency and wthe relations:
ysically significant, the value of ν (and therefor physical significance. Why?
on
2/3)e ;
e and m
very low
e out withrgy
avelength of
e, the value
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Quarks inside protons and neutrons carry fractional charges. This is because nuclear forceincreases extremely if they are pulled apart. Therefore, fractional charges may exist innature; observable charges are still the integral multiple of an electrical charge.
The basic relations for electric field and magnetic field are
.
These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius),and B (magnetic field). These relations give the value of velocity of an electron as
and
It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio e/m.
At atmospheric pressure, the ions of gases have no chance of reaching their respectiveelectrons because of collision and recombination with other gas molecules. Hence, gasesare insulators at atmospheric pressure. At low pressures, ions have a chance of reachingtheir respective electrodes and constitute a current. Hence, they conduct electricity atthese pressures.
The work function of a metal is the minimum energy required for a conduction electron to
get out of the metal surface. All the electrons in an atom do not have the same energylevel. When a ray having some photon energy is incident on a metal surface, the electronscome out from different levels with different energies. Hence, these emitted electronsshow different energy distributions.
The absolute value of energy of a particle is arbitrary within the additive constant. Hence,wavelength ( λ) is significant, but the frequency (ν) associated with an electron has nodirect physical significance.
Therefore, the product νλ(phase speed)has no physical significance.
Group speed is given as:
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This quantity has a physical eaning.