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Dual nature of radiation. Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; particle nature of light. Matter waves-wave nature of particle, de Broglie relation. Davisson-Germer experiment. Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars; isotones. Radioactivity- alpha, beta and gamma particles/rays and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number, nuclear fission and fusion.
PHOTOELECTRIC EFFECT
When electromagnetic radiations of suitable wavelength are incident on a metallic surface then electrons are emitted, this phenomenon is called photo electric effect.
G
Photoelectron
The electron emitted in photoelectric effect is called photoelectron.
Photoelectric Current
If current passes through the circuit in photoelectric effect then the current is called photoelectric current.
Work Function
The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denoted by ϕ or W. It is the minimum for Cesium. It is relatively less for alkali metals.
Work Functions of some Photosensitive Metals
Metal Work function (eV) Metal^
Work function (eV) Cesium 1.9 Calcium 3. Potassium 2.2 Copper 4. Sodium 2.3 Silver 4. Lithium 2.5 Platinum 5.
To produce photo electric effect only metal and light is necessary but for observing it the circuit is completed. Fig. 19.1 shows an arrangement used to study the photo- electric effect.
1 2
C A
I Intensity v Frequency
Rheostat
Cell, few volts
A
V
Fig. 19. Here the plate (1) is called emitter or cathode and other plate (2) is called collector or anode.
Chapter Highlights
Modern Physics
CHAPTER
19.2 Chapter 19
Saturation Current
When all the photo electrons emitted by cathode reach the anode then current flowing in the circuit at that instant is known as saturated current, this is the maximum value of photoelectric current.
Stopping Potential
Minimum magnitude of negative potential of anode with respect to cathode for which current is zero is called stop- ping potential. This is also known as cut-off voltage. This voltage is independent of intensity.
Retarding Potential
Negative potential of anode with respect to cathode which is less than stopping potential is called retarding potential.
OBSERVATIONS: (MADE BY EINSTEIN)
A graph between intensity of light and photoelectric cur- rent is found to be a straight line as shown in Fig. 19.2. Photoelectric current is directly proportional to the inten- sity of incident radiation. In this experiment the frequency and retarding potential are kept constant.
Photocurrent
O Intensity of light
Fig. 19.
A graph between photoelectric current and potential dif- ference between cathode and anode is found as shown in Fig. 19.3.
Saturation current
IS 1 v I 1
I 2 > I 1 IS 2 v
IP
Fig. 19.
In case of saturation current, rate of emission of photoelec- trons = rate of flow of photoelectrons, here, vs → stopping potential and it is a positive quantity
Electrons emitted from surface of metal have differ- ent energies. Maximum kinetic energy of photoelectron on the cathode = eVs KE max = eVs Whenever photoelectric effect takes place, electrons are ejected out with kinetic energies ranging from 0 to KE max i.e. 0 ≤ KEC ≤ eVs The energy distribution of photoelectron is shown in Fig. 19.4.
No. of Photoelectrons O Kinetic energy eVS
Fig. 19. If intensity is increased (keeping the frequency constant) then saturation current is increased by same factor by which intensity increases. Stopping potential is same, so maximum value of kinetic energy is not affected. If light of different frequencies is used then obtained plots are shown in Fig. 19.5. IP
v 3 > v 2 v 2 > v 1 v 1
- VS 3 – VS 2 – VS 1 VA – VC
Fig. 19. It is clear from graph, as n increases, stopping potential increases, it means maximum value of kinetic energy increases. Graphs between maximum kinetic energy of elec- trons ejected from different metals and frequency of light used are found to be straight lines of same slope as shown in Fig. 19.
m 1
vth 1 vth 2 vth 3 v
m 2 m 3
θ θ θ
kmax For metal’s
Fig. 19.
19.4 Chapter 19
Now, quantum theory solves these problems in pro- viding the correct interpretation of the photoelectric effect.
PLANCK’S QUANTUM THEORY
The light energy from any source is always an integral multiple of a smaller energy value called quantum of light. Hence, energy Q = NE ,
where E = h n
and N (number of photons) = 1,2,3,....
Here energy is quantized. h n is the quantum of energy, it is a packet of energy called as photon.
E = h n = hc λ
and hc ; 12400 eV Å
EINSTEIN’S PHOTON THEORY
In 1905 Einstein made a remarkable assumption about the nature of light; namely, that, under some circumstances, it behaves as if its energy is concentrated into localized bun- dles, later called photons. The energy E of a single photon is given by
E = h n,
If we apply Einstein’s photon concept to the photoelectric effect, we can write
h n = W + K max, (energy conservation)
Equation says that a single photon carries an energy h n into the surface where it is absorbed by a single electron. Part of this energy W (called the work function of the emitting surface) is used in causing the electron to escape from the metal surface. The excess energy ( h n – W ) becomes the electron kinetic energy; if the electron does not lose energy by internal collisions as it escapes from the metal, it will still have this much kinetic energy after it emerges. Thus K max represents the maximum kinetic energy that the pho- toelectron can have outside the surface. There is complete agreement of the photon theory with experiment.
Now IA = Nh n
⇒ N =
IA
h ν
= number of photons incident per unit time on an area A when light of intensity I is incident normally.
If we double the light intensity, we double the number of photons and thus double the photoelectric current; we do
not change the energy of the individual photons or the nature of the individual photoelectric processes. The second objection (the frequency problem) is met if K max equals zero, we have h nth = W , Which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear as kinetic energy. If n is reduced below nth, h n will be smaller than W and the individual photons, no matter how many of them there are (that is, no matter how intense the illumination), will not have enough energy to eject photoelectrons. The third objection (the time delay problem) follows from the photon theory because the required energy is sup- plied in a concentrated bundle. It is not spread uniformly over the beam cross section as in the wave theory. Hence Einstein’s equation for photoelectric effect is given by h n = h nth + K max K max = hc λ
hc λ th
SOLVED EXAMPLES
1. In an experiment on photo electric emission, following observations were made, (i) Wavelength of the incident light = 1.98 × 10 –7^ m; (ii) Stopping potential = 2.5 V Find: (A) Kinetic energy of photoelectrons with maximum speed. (B) Work function and (C) Threshold frequency. Solution: (A) Since vs = 2.5 V, K max = eVs so, K max = 2.5 eV (B) Energy of incident photon
E =
eV = 6.26 eV W = E – K max = 3.76 eV
(C) h nth = W = 3.76 × 1.6 × 10 –19^ J
\ nth =
19 34
× ×
×
≈ ×
− − Hz^.
2. A beam of light consists of four wavelength 4000 Å, 4800 Å, 6000 Å and 7000 Å, each of intensity 1.5 × 10 –3^ Wm–2. The beam falls normally on an area 10 –4^ m^2 of a clean metallic surface of work function 1.9 eV. Assuming no loss of light energy (i.e. each capable photon emits one electron) calculate the num- ber of photoelectrons liberated per second.
Modern Physics 19.
Solution:
E 1 =
= 3.1 eV
E 2 =
= 2.58 eV
E 3 =
= 2.06 eV
and E 4 =
= 1.77 eV
Therefore, light of wavelengths 4000 Å, 4800 Å and 6000 Å can only emit photoelectrons. \ Number of photoelectrons emitted per second = Number of photons incident per second)
=
I A
E
1 1 1
I A
E
2 2 2
I A
E
3 3 3
= I A
E 1 E 2 E 3
..^.^.
3 4 19
×
×
⎝⎜^
− − −
= 1.12 × 1012.
3. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 Å, no photoelectrons are emitted from the sur- face. With an unknown wavelength, stopping potential is 3 V. Find the unknown wavelength.
Solution: using equation of photoelectric effect K max = E – W ( K max = eV s )
\ 3 eV =
λ
λ
or l = 2262 Å.
4. Illuminating the surface of a certain metal alternately with light of wavelengths l 1 = 0.35 mm and l 2 = 0. mm, it was found that the corresponding maximum velocities of photo electrons have a ratio h = 2. Find the work function of that metal.
Solution: Using equation for two wavelengths 1 2 1
2 1
mv hc = − W λ
2 2
mv
hc = − W λ
Dividing Equation (1) with Equation (2), with v 1 = 2 v 2 ,
we have 4 =
hc W
hc W
λ
λ
1
2
3 W = 4
hc λ 2
⎠⎟^
hc λ 1
×
= 5.64 eV.
5. Light described at a place by the equation E = (100V/m) [sin (5 × 1015 s–1) t + sin (8 × 1015 s–1) t ] falls on a metal surface having work function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons.
Solution: The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy. This larger frequency is
ν
ω π π
× −
(^15) s 1
The maximum kinetic energy of the photoelectrons is K max = h u – W
= (4.14 × 10 –15^ eV-s) ×
15 ⎛ × 1
⎝
− π
s – 2.0 eV
= 5.27 eV – 2.0 eV = 3.27 eV.
6. Find the momentum of a 12.0 MeV photon.
Solution:
p =
E
c
= 12 MeV/c.
7. Monochromatic light of wavelength 3000 Å is incident normally on a surface of area 4 cm^2. If the intensity of the light is 15 × 10 –2^ W/m^2 , determine the rate at which photons strike the surface.
Solution: Rate at which photons strike the surface
IA
hc / λ
5 19
×
×
− −
J/s
. J/photon
= 9.05 × 1013 photon/s.
Modern Physics 19.
force due to absorbed photon ( Fa ) =
IA
hc
λ (1 – r ) h λ
IA
c
(1 – r ) (downward)
Force due to reflected photon ( Fr ) =
IA
hc
λ (^). r 2 h λ
2 IA
c
λ (downward)
Total force = Fa + Fr (downward)
=
IA
c
(1 – r ) + 2 IAr c
=
IA
c
(1 + r )
Now pressure P = (1 + r ) ×
A
I
c
(1 + r ).
SOLVED EXAMPLES
10. A plate of mass 10 gm is in equilibrium in air due to the force exerted by light beam on plate. Calculate power of beam. Assume plate is perfectly absorbing.
Solution: Since plate is in air, so gravitational force will act on this F gravitational = mg (downward)
= 10 × 10 –3^ × 10
= 10 –1^ N for equilibrium force exerted by light beam should be equal to F gravitational F photon = F gravitational Let power of light beam be P
\ F photon =
P
c
⇒
P
c
P = 3.0 × 10 8 × 10 –
P = 3 × 107 W.
11. Calculate force exerted by light beam if light is inci- dent on surface at an angle q as shown in Fig. 19.8. Consider all cases.
θ
Fig. 19.
Solution: Case: I a = 1, r = 0 initial momentum of photon (in downward direction at
an angle q with vertical) =
h λ
[ θ ]
final momentum of photon = 0 change in momentum (in upward direction at an
angle q with vertical) =
h λ
[^ θ^ ]
energy incident per unit time = IA cos q Intensity = power per unit normal area
I =
P
A cos θ
P = IA cos q
Number of photons incident per unit time =
IA
hc
cos .
θ λ
total change in momentum per unit time (in upward direction at an angle q with vertical)
=
IA
hc
cos θ λ⋅ .
h λ
IA
c
cos θ [^ θ^ ]
Force (F) = total change in momentum per unit time
F =
IA
c
cos θ (direction^ θ^ on photon and^ θ^ on the plate) Pressure = normal force per unit Area
Pressure =
F
A
cos θ P =
IA
cA
cos^2 θ
I
c
cos^2 q
Case: II When r = 1, a = 0 \ change in momentum of one photon
=
2 h λ
cos q (upward)
h λ sin^ θ
θ θ
h λ sin^ θ
h λ cos θ
h λ cos^ θ
19.8 Chapter 19
Number of photons incident per unit time
= Energy incident per unit time h ν
=
IA
hc
cos θ λ⋅
\ total change in momentum per unit time
IA
hc
cos θ λ⋅ × 2 h λ
cos q =
2 IA^2
c
cos θ (upward)
\ force on the plate =
2 IA^2
c
cos θ (downward)
Pressure =
2 IA^2
cA
cos θ
P =
2 I^2
c
cos θ
Case: III 0 < r < 1, a + r = 1 change in momentum of photon when it is reflected = 2 h λ
cos q (downward)
change in momentum of photon when it is absorbed
h λ
(in the opposite direction of incident beam)
energy incident per unit time = IA cos q
Number of photons incident per unit time =
IA
hc
cos θ λ⋅
Number of reflected photon ( nr ) = IA r hc
cos θ λ⋅
Number of absorbed photon ( nQ ) =
IA
hc
cos θ λ⋅ (1 – r )
Force on plate due to absorbed photons Fa = na ⋅ D Pa
=
IA
hc
cos θ λ⋅ (1 – r ) h λ
=
IA
c
cos θ (1 – r )
(at an angle q with vertical^ θ^ )
Force on plate due to reflected photons Fr = nr D Pr
=
IA
hc
cos θ λ⋅ × 2 h λ
cos θ(vertically downward)
IA
c
r cos^2 2 θ ⋅ Now resultant force is given by
FR = Fr^2^ + F (^) a^2^ + 2 F Fa r cos θ
IA
c
cos θ ( 1 − r ) 2 + ( 2 r ) 2 cos 2 θ+ 4 r r ( − 1 ) cos^2 θ
and, pressure
P =
F F
A
a cos^ θ^ + r
IA r cA
cos θ( 1 − ) cosθ
IA r cA
cos^2 θ ⋅ 2
I
c
cos^2 θ (1 – r ) +
I
c
cos^2 θ 2 r
I
c
cos^2 θ (1 + r ).
DE-BROGLIE WAVELENGTH OF
MATTER WAVE
A photon of frequency n and wavelength l has energy.
E h
hc = ν = λ By Einstein’s energy mass relation, E = mc^2 the equivalent mass m of the photon is given by,
m
E
c
h c
h c
ν λ
or λ λ
h c
or l =
h p Here p is the momentum of photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength l (called de-Broglie wavelength and the wave is called matter wave) given by, λ = =
h mv
h p
where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation,
p = 2 Km
and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these rela- tions Equation (19.2), can be written as,
λ = = = =
h mv
h p
h Km
h 2 2 qVm (de-Broglie wavelength) (19.2)
19.10 Chapter 19
as given by above formula. If this happened the radius of the orbit would decrease and the electron would spiral into the nucleus in a fraction of second. But atoms do not col- lapse. In 1913 an effort was made by Neils Bohr to over- come this paradox.
Regarding Explanation of Line Spectrum
In Rutherford’s model, due to continuously changing radii of the circular orbits of electrons, the frequency of revolu- tion of the electrons must be changing. As a result, elec- trons will radiate electromagnetic waves of all frequencies, i.e., the spectrum of these waves will be ‘continuous’ in nature. But experimentally the atomic spectra are not con- tinuous. Instead they are line spectra.
THE BOHR’S ATOMIC MODEL
In 1913, Prof. Niel Bohr removed the difficulties of Rutherford’s atomic model by the application of Planck’s quantum theory. For this he proposed the following postulates
1. An electron moves only in certain circular orbits, called stationary orbits. In stationary orbits electron does not emit radiation, contrary to the predictions of classical electromagnetic theory. 2. According to Bohr, there is a definite energy associated with each stable orbit and atom radiates energy only when it makes a transition from one of these orbits to another. If the energy of electron in the higher orbit is E 2 and that in the lower orbit be E 1 , then the frequency n of the radiated waves is given by h n = E 2 – E 1
or n =
E E
h
3. Bohr found that the magnitude of the electron’s angu- lar momentum is quantized, and this magnitude for the electron must be integral multiple of h 2 π . The magni-
tude of the angular momentum is L = mvr for a particle with mass m moving with speed v in a circle of radius r. So, according to Bohr’s postulate,
mvr nh = 2 π
( n = 1, 2, 3....) (19.4)
Each value of n corresponds to a permitted value of the orbit radius, which we will denote by rn. The value of n for each orbit is called principal quantum number for the orbit. Thus,
mvnrn = mvr nh = 2 π
According to Newton’s second law a radially inward cen-
tripetal force of magnitude F = mv rn
2 is needed by the elec-
tron which is being provided by the electrical attraction between the positive proton and the negative electron.
Thus, mv r
e r
n n (^) n
2
0
2 2
πε
Solving Equation (19.5) and (19.6), we get
r n h me n =
ε π
0
2 2 2 (19.7)
and v e n (^) nh
2
2 ε (^0)
The smallest orbit radius corresponds to n = 1. We’ll denote this minimum radius, called the Bohr radius as a 0. Thus, ν Substituting values of e 0 , h , p , m and e , we get a 0 = 0.529 × 10 –10, m = 0.529 Å (19.9) Equation (19.7), in terms of a 0 can be written as, rn = n^2 a 0 or rn ∝ n^2 (19.10) Similarly, substituting values of e , e 0 and h with n = 1 in Equation (19.8), we get v 1 = 2.19 × 106 m/s (9.11) This is the greatest possible speed of the electron in the hydrogen atom, which is approximately equal to c/137, where c is the speed of light in vacuum. Equation (19.9), in terms of v 1 can be written as,
vn = v n
(^1) or v n ∝^
n
Energy Levels
Kinetic and potential energies Kn and Un in n th^ orbit are given by Kn =
mvn^2 = me n h
4
0 8 ε^2 2
and Un = –
4 πε 0
e rn
2 = – me n h
4
0 4 ε^2 2
(assuming infinity as a zero potential energy level) The total energy En is the sum of the kinetic and potential energies.
so, En = Kn + Un = – me n h
4
0 8 ε^2 2
Modern Physics 19.
Substituting values of m , e , e 0 and h with n = 1, we get the least energy of the atom in first orbit, which is –13.6 eV. Hence,
E 1 = – 13.6 eV (19.12)
and En =
E
n
1 2 =^ –^
2
n
eV (19.13)
Substituting n = 2, 3, 4, ...., etc., we get energies of atom in different orbits.
E 2 = – 3.40 eV, E 3 = – 1.51 eV, .... E ∞ = 0
Hydrogen Like Atoms
The Bohr model of hydrogen can be extended to hydrogen like atoms, i.e., one electron atoms, the nuclear charge is +ze, where z is the atomic number, equal to the number of protons in the nucleus. The effect in the previous analysis is to replace e^2 every where by ze^2. Thus, the equations for, rn , vn and En are altered as under
rn =
ε 0 2 2 2
n h nmze
n z
2 a 0 or rn ∝
n z
2
where a 0 = 0.529 Å (radius of first orbit of H )
vn =
ze nh
2
2 ε (^0)
z n
v 1 or vn ∝
z n
where v 1 = 2.19 × 106 m/s (speed of electron in first orbit of H )
En = –
mz e n h
2 4
0 8 ε 2 2 2 =^
z n
2 2 E^1 or^ En^ ∝^
z n
2 2
where E 1 = –13.60 eV (energy of atom in first orbit of H )
Definitions Valid for Single Electron System
Ground State
Lowest energy state of any atom or ion is called ground state of the atom. Ground state energy of H atom = –13.6 eV Ground state energy of He+^ Ion = –54.4 eV Ground state energy of Li++^ Ion = –122.4 eV
Excited State
State of atom other than the ground state are called its excited states. n = 2 first excited state n = 3 second excited state
n = 4 third excited state n = n 0 + 1 n 0 th^ excited state
Ionization Energy (I****. E.) Minimum energy required to move an electron from ground state to n = ∞ is called ionization energy of the atom or ion Ionization energy of H atom = 13.6 eV Ionization energy of He+^ Ion = 54.4 eV Ionization energy of Li++^ Ion = 122.4 eV
Ionization Potential (I.P.) Potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to ionization energy of the atom is called ionization potential of the atom. I.P of H atom = 13.6 V I.P. of He+^ Ion = 54.4 V
Excitation Energy Energy required to move an electron from ground state of the atom to any other exited state of the atom is called excitation energy of that state. Energy in ground state of H atom = –13.6 eV Energy in first excited state of H atom = –3.4 eV Ist^ excitation energy = 10.2 eV.
Excitation Potential Potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to excitation energy of any state is called excitation potential of that state. Ist^ excitation energy = 10.2 eV. Ist^ excitation potential = 10.2 V.
Binding Energy or Separation Energy Energy required to move an electron from any state to n = ∞ is called binding energy of that state or energy released dur- ing formation of an H-like atom/ion from n = ∞ to some particular n is called binding energy of that state. Binding energy of ground state of H atom = 13.6 eV
SOLVED EXAMPLE
16. First excitation potential of a hypothetical hydrogen like atom is 15 V Find third excitation potential of the atom.
Solution: Let energy of ground state = E 0
E 0 = – 13.6 Z^2 eV and En =
E
n
0 2
Modern Physics 19.
Series Limit Line of any group having maximum energy of photon and minimum wavelength of that group is called series limit.
Lymen series
Paschen series
Pfund series
Balmer series
Brackett series –0.28 eV –0.38 eV –0.54 eV –0.85 eV –1.51 eV
–3.40 eV
n = 1 –13.6 eV
n = 2
n = 3
n = 4
n = 5 n = 6
n = 7
For the Lymen series nf = 1, for Balmer series nf = 2 and so on.
Wavelength of Photon Emitted in
De-excitation
According to Bohr when an atom makes a transition from higher energy level to a lower level it emits a photon with energy equal to the energy difference between the initial and final levels. If Ei is the initial energy of the atom before such a transition, Ef is its final energy after the transition, and the photon’s energy is h n = hc λ
, then conservation of energy gives,
h n = hc λ
= Ei – Ef (energy of emitted photon)
By 1913, the spectrum of hydrogen had been studied inten- sively. The visible line with longest wavelength, or lowest frequency is in the red and is called H a , the next line, in the blue-green is called H b and so on. In 1885, Johann Balmer, a Swiss teacher found a for- mula that gives the wave lengths of these lines. This is now called the Balmer series. The Balmer’s formula is, 1 1 2
λ =^2 −^2
⎝⎜^
R
n
Here, n = 3, 4, 5 ...., etc. R = Rydberg constant = 1.097 × 107 m– and l is the wavelength of light/photon emitted dur- ing transition, For n = 3, we obtain the wavelength of H a line. Similarly, for n = 4, we obtain the wavelength of H b line. For n = ∞, the smallest wavelength (= 3646 Å) of this series is obtained. Using the relation, E = hc λ
we can find the photon energies corresponding to the wavelength of the Balmer series.
E
hc hcR n
Rhc Rhc n
⎝⎜^
⎠⎟^
λ
This formula suggests that,
En = –
Rhc n^2
, n = 1, 2, 3.....
The wavelengths corresponding to other spectral series (Lymen, Paschen, (etc.) can be represented by formula sim- ilar to Balmer’s formula.
Lymen Series:
λ =^2 −^2
⎝⎜^
R
n
, n = 2, 3, 4.....
Paschen Series:
λ =^2 −^2
⎝⎜^
R
n
, n = 4, 5, 6.....
Brackett Series:
λ =^2 −^2
⎝⎜^
R
n
, n = 5, 6, 7.....
Pfund Series:
m
, n = 6, 7, 8
The Lymen series is in the ultraviolet, and the Paschen. Brackett and Pfund series are in the infrared region.
SOLVED EXAMPLES
17. Calculate (A) the wavelength and (B) the frequency of the H b line of the Balmer series for hydrogen.
Solution: (A) H b line of Balmer series corresponds to the transi- tion from n = 4 to n = 2 level. The corresponding wavelength for H b line is, 1 1 097 10
7 λ =^ ×^2 −^2
⎝⎜^
= 0.2056 × 107 l = 4.9 × 10 –7^ m
(B) n = r 0 =
8 7
×
× −^
= 6.12 × 1014 Hz.
18. Find the largest and shortest wavelengths in the Lymen series for hydrogen. In what region of the electromag- netic spectrum does each series lie?
Solution: The transition equation for Lymen series is given by,
1 1 1
λ =^2 −^2
R ⎥
( ) n
n = 2, 3, ......
19.14 Chapter 19
for largest wavelength, n = 2 1 1 097 10
7 λmax
=. × −
⎝⎜^
⎠⎟^
= 0.823 × 107
\ lmax = 1.2154 × 10 –7^ m = 1215 Å The shortest wavelength corresponds to n = ∞
\
λmax
=. × −
⎝⎜^
or lmin = 0.911 × 10 –7^ m = 911 Å Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum.
19. How many different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n?
Solution: From the nth state, the atom may go to ( n – 1)th state, ...., 2nd state or 1st state. So there are ( n – 1) possi- ble transitions starting from the nth state. The atoms reaching ( n – 1)th state may make ( n – 2) different transitions. Similarly for other lower states. The total number of possible transitions is ( n – 1) + ( n – 2) + ( n – 3) +............2 + 1
=
n n ( − 1 ) 2
(Remember).
20. (A) Find the wavelength of the radiation required to excite the electron in Li++^ from the first to the third Bohr orbit. (B) How many spectral lines are observed in the emis- sion spectrum of the above excited system? Solution: (A) The energy in the first orbit = E 1 = Z^2 E 0 where E 0 = – 13.6 eV is the energy of a hydrogen atom in ground state thus for Li++,
E 1 = 9 E 0 = 9 × (– 13.6 eV) = – 122.4 eV
The energy in the third orbit is
E 3 =
E
n
1^ E
2
1 9
= = – 13.6 eV
Thus, E 3 – E 1 = 8 × 13.6 eV = 108.8 eV.
Energy required to excite Li++^ from the first orbit to the third orbit is given by E 3 – E 1 = 8 × 13.6 eV = 108.8 eV.
The wavelength of radiation required to excite Li++ from the first orbit to the third orbit is given by hc E E λ
or, λ = −
hc E (^) 3 E 1
eV nm eV
nm
(B) The spectral lines emitted are due to the transi- tions n = 3 → n = 2, n = 3 → n = 1 and n = 2 → n = 1. Thus, there will be three spectral lines in the spectrum.
21. Find the kinetic energy potential energy and total energy in first and second orbit of hydrogen atom if potential energy in first orbit is taken to be zero.
Solution:
E 1 = – 13.60 eV K 1 = – E 1 = 13.60 eV
U 1 = 2E 1 = –27.20 eV
E 2 =
E 1
( ) 22 =^ – 3.40 eV K 2 = 3.40 eV and U 2 = – 6.80 eV Now U 1 = 0, i.e., potential energy has been increased by 27.20 eV while kinetic energy will remain unchanged. So values of kinetic energy, potential energy and total energy in first orbit are 13.60 eV, 0, 13.60 respectively and for second orbit these values are 3.40 eV, 20.40 eV and 23.80 eV.
22. A lithium atom has three electrons, Assume the fol- lowing simple picture of the atom. Two electrons move close to the nucleus making up a spherical cloud around it and the third moves outside this cloud in a circular orbit. Bohr’s model can be used for the motion of this third electron but n = 1 states are not available to it. Calculate the ionization energy of lithium in ground state using the above picture.
Solution: In this picture, the third electron moves in the field of a total charge + 3 e – 2 e = + e. Thus, the energies are the same as that of hydrogen atoms. The lowest energy is:
E 2 = Ec =
. eV = – 3.4 eV
19.16 Chapter 19
For the given particle,
E 4 =
× =
2 2
eV
= –1360 eV
and E 2 =
2 2 ×^100 =^ – 5440 eV
E = E 4 – E 2 = 4080 eV
\ l (in Å) =
= 3.0 Å.
26. A gas of hydrogen like atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon. (A) Determine the initial state of the gas atoms. (B) Identify the gas atoms. (C) Find the minimum wavelength of the emitted radiations. (D) Find the ionization energy and the respective wavelength for the gas atoms. Solution:
(A) n n ( − ) =
\ n = 3 i.e., after excitation atom jumps to second excited state. Hence nf = 3. So ni can be 1 or 2 If ni = 1 then energy emitted is either equal to, greater than or less than the energy absorbed. Hence the emitted wavelength is either equal to, less than or greater than the absorbed wavelength. Hence ni ≠ 1. If ni = 2, then Ee ≥ Ea. Hence l e ≤ l 0 (B) E 3 – E 2 = 68 eV
\ (13.6) ( Z^2 )
⎝⎜^
⎠⎟^
\ Z = 6
(C) lmin =
E 3 − E 1
(. ) ( )^2 ⎛ −
⎝⎜^
(D) Ionization energy = (13.6) (6)^2 = 489.6 eV
l =
= 25.33 Å.
27. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr’s postulate regarding the quantization of angular momentum holds good for this electron, find (A) The allowed values of the radius r of the orbit. (B) The kinetic energy of the electron in orbit. (C) The potential energy of interaction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B. (D) The total energy of the allowed energy levels. Solution: (A) Radius of circular path
r = mv Be
mvr = nh 2 π
Solving these two equations, we get
r = nh 2 π Be
and v = nhBe 2 π m^2
(B) K =
mv^2 = nhBe 4 π m
(C) M = iA = e T
⎝⎜^
⎠⎟^
(^ p^ r^^2 )^ =^
evr 2
= e nh Be
nhBe 2 2 π (^2) π m^2
= λ π
hc 4 m Now potential energy U = – M. B
= λ π
heB 4 m
(D) E = U + K = nheB 2 π m
28. Determine the wavelength of the second line of the Paschen series for hydrogen.
Solution: 1 λ
= (1.097 × 10 –3^ Å–1)
⎝⎜^
or l = 12,820 Å.
Modern Physics 19.
29. How many different photons can be emitted by hydro- gen atoms that undergo transitions to the ground state from the n = 5 state?
Solution: 10 photons.
30. An electron rotates in a circle around a nucleus with positive charge Ze. How is the electrons’ velocity related to the radius of its orbit?
Solution:
v kZe mr
2 .
EFFECT OF NUCLEUS MOTION ON
ENERGY OF ATOM
Let both the nucleus of mass M, charge Ze and electron of mass m, and charge e revolve about their centre of mass (CM) with same angular velocity ( w) but different linear speeds. Let r 1 and r 2 be the distance of CM from nucleus and electron. Their angular velocity should be same then only their separation will remain unchanged in an energy level.
r (^2) m CM M^ r^1
Let r be the distance between the nucleus and the electron. Then
Mr 1 = mr 2 r 1 + r 2 = r
\ r 1 = mr M + m
and r 2 = Mr M + m
Centripetal force to the electron is provided by the electro- static force. So,
mr 2 w^2 =
2
πε^2
Ze r
or m Mr M + m
⎝⎜^
w^2 =
4 πε (^0)
. Ze r
2 2
or Mm M + m
⎝⎜^
⎠⎟^
r^3 w^2 = Ze^2 4 πε 0
or m r^3 w^2 = e^2 4 πε 0
where Mm M + m
= m
Moment of inertia of atom about CM,
I = Mr 12 + mr 22 = Mm M + m
⎝⎜^
⎠⎟^
r^2 =^ m r^2
According to Bohr’s theory, nh 2 π
= I w
or m r^2 w = nh 2 π
Solving above equations for r, we get
r = ε πμ
0
2 2 2
n h e Z
and r = (0.529 Å) n Z
(^2) m ⋅ μ
Further electrical potential energy of the system,
U
Ze r
4 πε (^0)
U =
− Z e n h
2 4
0
μ ε
and kinetic energy,
K =
I w^2 =
μ r^2^ ω^2 and K = μN^2
and K =
mv^2
v -speed of electron with respect to nucleus. ( v = r w)
here w^2 = Ze r
2
0 4 πε μ^3
\ K =
Ze r
Z e n h
2
0
2 4
0 8 πε 8 2 2 2
μ πε
\ Total energy of the system En = K + U
En = – μ ε
e n h
4
0
this expression can also be written as
En = – (13.6 eV)
Z
n m
2 2 ⋅
⎝⎜^
μ
The expression for En without considering the motion of
proton is En = – me n h
4
0 8 ε 2 2 2 , i.e.,^ m^ is replaced by^ m^ while considering the motion of nucleus.
Modern Physics 19.
He+^ ion. Find minimum value of K so that there can be an inelastic collision between these two particles.
Solution: Here the loss during the collision can only be used to excite the atoms or electrons. So according to quantum mechanics loss = {0, 40.8 eV, 48.3 eV, ......, 54.4 eV} (1)
En = –13.
Z
n
2 2 eV Now according to newtonian mechanics
n
m K (^) He+
4 m
Minimum loss = 0 maximum loss will be for perfectly inelastic collision. let v 0 be the initial speed of neutron and vf be the final common speed. so by momentum conservation mv 0 = mvf + 4 mvf vf =
v 0 5
where m = mass of Neutron \ mass of He+^ ion = 4 m so final kinetic energy of system
KE =
m v (^) f^2 +
4 m v^2 f
0
2 0 ⋅ ( m) ⋅ v^ = .( mv^2 )= K
Maximum loss = K –
K
K
so loss will be 0
⎡ K
⎣^ ⎢^
⎦^ ⎥^
For inelastic collision there should be at least one com- mon value other than zero in set (1) and (2)
\
K
> 40.8 eV
K > 51 eV Minimum value of K = 51 eV.
35. In previous question, find minimum value of K so that all types of collision is possible.
Solution:
K =
× 12.09 ⇒ K = 60.45 eV.
36. A H-atom in ground state is moving with initial kinetic energy K. It collides head on with a He+^ ion in ground state kept at rest but free to move. Find minimum value of K so that both the particles can excite to their first excited state.
Solution: 4 5
K
= 51 eV K =
×
eV K = 63.75 eV.
37. How many head-on, elastic collisions must a neutron have with deuterium nucleus to reduce its energy from 1 MeV to 0.025 eV.
Solution: Let mass of neutron = m and mass of deuterium = 2 m initial kinetic energy of neutron = K 0 Let after first collision kinetic energy of neutron and deuterium be K 1 and K 2. Using C.O.L.M. along direction of motion
2 mK 0 = 2 mK 1 + 4 mK 2
velocity of separation = velocity of approach
4 2
mK 2 m
2 mK 1 m
2 mK 0 m Solving Equation on (1) and (2) we get K 1 =
K 0
Loss in kinetic energy after first collision D K 1 = K 0 – K 1
D K 1 =
K 0 (1)
After second collision
D K 2 =
K 1 =
K 0
\ Total energy loss D K = D K 1 + D K 2 + ..... + D Kn
As, D K =
K 0 +
K 0 + .......... +
n K^0
D K =
K 0 (1 +
9 n −^1
Δ K
K
n
0
9 n
Here, K 0 = 106 eV, D K = (10^6 – 0.025) eV
19.20 Chapter 19
\
9 n^
K K
K
0 0
or 9 n^ = 4 × 107 Taking log both sides and solving, we get n = 8.
38. A neutron with energy of 4.6 MeV collides with pro- tons and is retarded. Assuming that upon each colli- sion neutron is deflected by 45º find the number of collisions which will reduce its energy to 0.23 eV.
Solution: Mass of neutron mass of proton = m
m K 0 Neutron
m
Proton
⇒
Neutron
Proton
45º
θº
K 2
K 1
x
y
From conservation of momentum in y -direction 2 mK 1 sin 45º 2 mK (^) 2 = sin q (1) In x -direction
2 mK (^) 0 – 2 mK 1 cos 45º = 2 mK 2 cos q (2)
Squaring and adding Equations (1) and (2), we have K 2 = K 1 + K 0 – 2 K K 0 1 (3) From conservation of energy K 2 = K 0 – K 1 (4) Solving Equations (3) and (4), we get
K 1 =
K 0
i.e., after each collision energy remains half. Therefore, after n collisions, Kn = K 0
⎝⎜^
n
\ 0.23 = (4.6 × 106 )
⎝⎜^
n 2
6 n (^) =. × . Taking log and solving, we get n ≈ 24.
Calculation of Recoil Speed of Atom on
Emission of a Photon
momentum of photon = mc = h λ (a) H atom in first excited state hc λ = 10.2 eV
Fixed
(b)
v H-atom
Free to move
h λ′
m - mass of atom According to momentum conservation
mv =
h λ ′
According to energy conservation 1 2
m ν^2 hc λ
= 10.2 eV
Since mass of atom is very large than photon
hence
m ν 2 can be neglected
hc λ ′ = 10.2 eV
h λ
c
eV
m n =
c
eV n =
cm
recoil speed of atom =
cm
X-RAYS
It was discovered by roentgen. The wavelength of x -rays is found between 0.1 Å to 10 Å. These rays are invisible to eye. They are electromagnetic waves and have speed c = 3 × 108 m/s in vacuum. Its photons have energy around 1000 times more than the visible light.
Rw mw IR v uv x γ
v Increases
When fast moving electrons having energy of order of sev- eral KeV strike the metallic target then x -rays are produced.
