Modern Physics | Physics Optional Notes for UPSC PDF Download

 Page 1


MODERN PHYSICS 
PHOTOELECTRIC EFFECT 
¦ PHOTON 
Quantum physics proposes that certain events can only be understood if we treat light as 
made up of particles. These particles of light are known as photons. 
A photon can be assumed as a packet of energy. 
Properties of photon: 
1. The speed of a photon in a vacuum is the same as the speed of light in a vacuum. 
2.  i.e. c=3×10
8
 m/s. It is independent of the choice of reference frame. 
3. Momentum of photon =
h
?? 
4. All photons of light of a given frequency (or wavelength have the same energy and 
momentum.) 
5. Photons are electrically neutral and are not deflected by electric and magnetic 
fields. 
6. In a photon particle collision. The total energy and momentum are conserved. 
However, no. of photons may not be conserved in a collision. Photon may be 
absorbed or a new photon may be created. 
The energy of a photon =
h?? ?? =h?? 
?? = Wavelength of light 
?? = Frequency of light 
C= Speed of light 
h= Planck's constant 
h=6.626×10
-34
Js 
P=
E
C
 
1?? m=10
-6
 m 
1 nm=10
-9
 m 
Page 2


MODERN PHYSICS 
PHOTOELECTRIC EFFECT 
¦ PHOTON 
Quantum physics proposes that certain events can only be understood if we treat light as 
made up of particles. These particles of light are known as photons. 
A photon can be assumed as a packet of energy. 
Properties of photon: 
1. The speed of a photon in a vacuum is the same as the speed of light in a vacuum. 
2.  i.e. c=3×10
8
 m/s. It is independent of the choice of reference frame. 
3. Momentum of photon =
h
?? 
4. All photons of light of a given frequency (or wavelength have the same energy and 
momentum.) 
5. Photons are electrically neutral and are not deflected by electric and magnetic 
fields. 
6. In a photon particle collision. The total energy and momentum are conserved. 
However, no. of photons may not be conserved in a collision. Photon may be 
absorbed or a new photon may be created. 
The energy of a photon =
h?? ?? =h?? 
?? = Wavelength of light 
?? = Frequency of light 
C= Speed of light 
h= Planck's constant 
h=6.626×10
-34
Js 
P=
E
C
 
1?? m=10
-6
 m 
1 nm=10
-9
 m 
1 Å=10
-10
 m 
1pm=10
-12
 m 
1fm=10
-15
 m 
1 nm=10 Å 
1 Å=0.1 nm 
1eV is the energy required by one electron to move through 1 V. 
1eV=1.6×10
-19
 J 
6.25×10
18
eV=1 J 
0.625×10
19
eV=1 J 
Example. h= eVs 
Solution: 1eV=1.6×10
-19
 J 
1 J=
1
1.6
×10
19
eV
? h =6.626×10
-34
Js
 =6.626×10
-34
×
1
1.6
×10
19
eVs
 =10
-15
×
6.626
1.6
eVs
 =4.14×10
-15
eVs
h =4.14×10
-15
eVs
 
Example. The energy of a photon in eV with ?? =621 nm 
Solution : ?? =621 nm=621×10
-9
 m 
E=
hc
?? =
4.14×10
-15
×3×10
8
621×10
-9
=
414
207
=2eV
hc=1242eVnm or 1240eVnm (depending upon calculation ) 
 
Example. How many photons are emitted per second by a 6-watt laser operating at 
663 mm ? 
Solution: ?? =6 watt 
1sec?6 J Energy 
E=n
hc
?? 
Page 3


MODERN PHYSICS 
PHOTOELECTRIC EFFECT 
¦ PHOTON 
Quantum physics proposes that certain events can only be understood if we treat light as 
made up of particles. These particles of light are known as photons. 
A photon can be assumed as a packet of energy. 
Properties of photon: 
1. The speed of a photon in a vacuum is the same as the speed of light in a vacuum. 
2.  i.e. c=3×10
8
 m/s. It is independent of the choice of reference frame. 
3. Momentum of photon =
h
?? 
4. All photons of light of a given frequency (or wavelength have the same energy and 
momentum.) 
5. Photons are electrically neutral and are not deflected by electric and magnetic 
fields. 
6. In a photon particle collision. The total energy and momentum are conserved. 
However, no. of photons may not be conserved in a collision. Photon may be 
absorbed or a new photon may be created. 
The energy of a photon =
h?? ?? =h?? 
?? = Wavelength of light 
?? = Frequency of light 
C= Speed of light 
h= Planck's constant 
h=6.626×10
-34
Js 
P=
E
C
 
1?? m=10
-6
 m 
1 nm=10
-9
 m 
1 Å=10
-10
 m 
1pm=10
-12
 m 
1fm=10
-15
 m 
1 nm=10 Å 
1 Å=0.1 nm 
1eV is the energy required by one electron to move through 1 V. 
1eV=1.6×10
-19
 J 
6.25×10
18
eV=1 J 
0.625×10
19
eV=1 J 
Example. h= eVs 
Solution: 1eV=1.6×10
-19
 J 
1 J=
1
1.6
×10
19
eV
? h =6.626×10
-34
Js
 =6.626×10
-34
×
1
1.6
×10
19
eVs
 =10
-15
×
6.626
1.6
eVs
 =4.14×10
-15
eVs
h =4.14×10
-15
eVs
 
Example. The energy of a photon in eV with ?? =621 nm 
Solution : ?? =621 nm=621×10
-9
 m 
E=
hc
?? =
4.14×10
-15
×3×10
8
621×10
-9
=
414
207
=2eV
hc=1242eVnm or 1240eVnm (depending upon calculation ) 
 
Example. How many photons are emitted per second by a 6-watt laser operating at 
663 mm ? 
Solution: ?? =6 watt 
1sec?6 J Energy 
E=n
hc
?? 
6=?? 6.626×10
-34
×3×10
8
Jsm/s
663×10
-9
 m
 
2=?? 6.663×10
-17
663
 
n=
2×10
2
10
-17
=2×10
19
 
Hence number of photons emitted per second =2×10
19
 
Example. Monochromatic light of frequency 
4
6.63
×10
14
 Hz is produced by source of 
power 2 mW . How many photons per second are emitted? 
Solution : ?? =
4
6.63
×10
14
 Hz 
?? =???? ??? =
?? ?? 
P=2×10
-3
 W 
1sec=2×10
-3
 J 
2×10
-3
=n
hc
?? ? 2×10
-3
=n×6.63×10
-34
×
4
6.63
×10
14
 
n=
1
2
×10
17
=5×10
16
 
n=5×10
16
 
Momentum of a photon : 
Example
?? =
?? ?? E.P. =
h
?? =
6.63×10
-34
3×10
8
+
4×10
14
6.63
=
4
3
×10
-28
=
2×10
-3
5×10
16
×3×10
8
=
4×10
-4
3×10
24
=
4
3
×10
-28
 
T. Source is of power =W. Find force exerted on wall if 
?? ?
?
?
 
(a) all photons are absorbed. 
(b) If all photons are reflected 
(c) 60% absorbed and 40% reflested. 
(d) All reflected 
Solution : (a) Let ?? be wavelength of light and ?? photons per sec are emitted. 
I sec? W Joule 
Page 4


MODERN PHYSICS 
PHOTOELECTRIC EFFECT 
¦ PHOTON 
Quantum physics proposes that certain events can only be understood if we treat light as 
made up of particles. These particles of light are known as photons. 
A photon can be assumed as a packet of energy. 
Properties of photon: 
1. The speed of a photon in a vacuum is the same as the speed of light in a vacuum. 
2.  i.e. c=3×10
8
 m/s. It is independent of the choice of reference frame. 
3. Momentum of photon =
h
?? 
4. All photons of light of a given frequency (or wavelength have the same energy and 
momentum.) 
5. Photons are electrically neutral and are not deflected by electric and magnetic 
fields. 
6. In a photon particle collision. The total energy and momentum are conserved. 
However, no. of photons may not be conserved in a collision. Photon may be 
absorbed or a new photon may be created. 
The energy of a photon =
h?? ?? =h?? 
?? = Wavelength of light 
?? = Frequency of light 
C= Speed of light 
h= Planck's constant 
h=6.626×10
-34
Js 
P=
E
C
 
1?? m=10
-6
 m 
1 nm=10
-9
 m 
1 Å=10
-10
 m 
1pm=10
-12
 m 
1fm=10
-15
 m 
1 nm=10 Å 
1 Å=0.1 nm 
1eV is the energy required by one electron to move through 1 V. 
1eV=1.6×10
-19
 J 
6.25×10
18
eV=1 J 
0.625×10
19
eV=1 J 
Example. h= eVs 
Solution: 1eV=1.6×10
-19
 J 
1 J=
1
1.6
×10
19
eV
? h =6.626×10
-34
Js
 =6.626×10
-34
×
1
1.6
×10
19
eVs
 =10
-15
×
6.626
1.6
eVs
 =4.14×10
-15
eVs
h =4.14×10
-15
eVs
 
Example. The energy of a photon in eV with ?? =621 nm 
Solution : ?? =621 nm=621×10
-9
 m 
E=
hc
?? =
4.14×10
-15
×3×10
8
621×10
-9
=
414
207
=2eV
hc=1242eVnm or 1240eVnm (depending upon calculation ) 
 
Example. How many photons are emitted per second by a 6-watt laser operating at 
663 mm ? 
Solution: ?? =6 watt 
1sec?6 J Energy 
E=n
hc
?? 
6=?? 6.626×10
-34
×3×10
8
Jsm/s
663×10
-9
 m
 
2=?? 6.663×10
-17
663
 
n=
2×10
2
10
-17
=2×10
19
 
Hence number of photons emitted per second =2×10
19
 
Example. Monochromatic light of frequency 
4
6.63
×10
14
 Hz is produced by source of 
power 2 mW . How many photons per second are emitted? 
Solution : ?? =
4
6.63
×10
14
 Hz 
?? =???? ??? =
?? ?? 
P=2×10
-3
 W 
1sec=2×10
-3
 J 
2×10
-3
=n
hc
?? ? 2×10
-3
=n×6.63×10
-34
×
4
6.63
×10
14
 
n=
1
2
×10
17
=5×10
16
 
n=5×10
16
 
Momentum of a photon : 
Example
?? =
?? ?? E.P. =
h
?? =
6.63×10
-34
3×10
8
+
4×10
14
6.63
=
4
3
×10
-28
=
2×10
-3
5×10
16
×3×10
8
=
4×10
-4
3×10
24
=
4
3
×10
-28
 
T. Source is of power =W. Find force exerted on wall if 
?? ?
?
?
 
(a) all photons are absorbed. 
(b) If all photons are reflected 
(c) 60% absorbed and 40% reflested. 
(d) All reflected 
Solution : (a) Let ?? be wavelength of light and ?? photons per sec are emitted. 
I sec? W Joule 
W=N(
hc
?? ) ? 
h
?? =
W
Nc
 
?? =?? |??? |=|0-
?? h
?? |=
?? ?? 
(b) ?? =|
-Nh
?? -
Nh
?? |=
2 W
c
 
(c) ?? =|
-40
100
Nh
?? -
Nh
?? |=
Nh
?? |0.4+01|?F=1.4
W
c
 
OR 
F=
60
100
W
c
-
40
100
(
2 W
c
) [
W
c
=100% absorbed ;
2 W
c
=100% reflected ] 
=
0.6 W
c
+
0.8 W
c
=1.4
W
c
 
(d) ?? =|-2
Nh
?? cos 60
°
|=
W
c
 
 
Work Function of a substance (?? ) : 
Minimum energy required to exit an electron from surface of the substance is called 
work function of the substance. 
Photoelectric Effect : 
Light is incident on metals surface with work function (?? ) . 
A photon may or may not collide with on electron. 
Consider a situation in which a photon collide with N electron and gives 100% of its 
energy to that electron now 
Page 5


MODERN PHYSICS 
PHOTOELECTRIC EFFECT 
¦ PHOTON 
Quantum physics proposes that certain events can only be understood if we treat light as 
made up of particles. These particles of light are known as photons. 
A photon can be assumed as a packet of energy. 
Properties of photon: 
1. The speed of a photon in a vacuum is the same as the speed of light in a vacuum. 
2.  i.e. c=3×10
8
 m/s. It is independent of the choice of reference frame. 
3. Momentum of photon =
h
?? 
4. All photons of light of a given frequency (or wavelength have the same energy and 
momentum.) 
5. Photons are electrically neutral and are not deflected by electric and magnetic 
fields. 
6. In a photon particle collision. The total energy and momentum are conserved. 
However, no. of photons may not be conserved in a collision. Photon may be 
absorbed or a new photon may be created. 
The energy of a photon =
h?? ?? =h?? 
?? = Wavelength of light 
?? = Frequency of light 
C= Speed of light 
h= Planck's constant 
h=6.626×10
-34
Js 
P=
E
C
 
1?? m=10
-6
 m 
1 nm=10
-9
 m 
1 Å=10
-10
 m 
1pm=10
-12
 m 
1fm=10
-15
 m 
1 nm=10 Å 
1 Å=0.1 nm 
1eV is the energy required by one electron to move through 1 V. 
1eV=1.6×10
-19
 J 
6.25×10
18
eV=1 J 
0.625×10
19
eV=1 J 
Example. h= eVs 
Solution: 1eV=1.6×10
-19
 J 
1 J=
1
1.6
×10
19
eV
? h =6.626×10
-34
Js
 =6.626×10
-34
×
1
1.6
×10
19
eVs
 =10
-15
×
6.626
1.6
eVs
 =4.14×10
-15
eVs
h =4.14×10
-15
eVs
 
Example. The energy of a photon in eV with ?? =621 nm 
Solution : ?? =621 nm=621×10
-9
 m 
E=
hc
?? =
4.14×10
-15
×3×10
8
621×10
-9
=
414
207
=2eV
hc=1242eVnm or 1240eVnm (depending upon calculation ) 
 
Example. How many photons are emitted per second by a 6-watt laser operating at 
663 mm ? 
Solution: ?? =6 watt 
1sec?6 J Energy 
E=n
hc
?? 
6=?? 6.626×10
-34
×3×10
8
Jsm/s
663×10
-9
 m
 
2=?? 6.663×10
-17
663
 
n=
2×10
2
10
-17
=2×10
19
 
Hence number of photons emitted per second =2×10
19
 
Example. Monochromatic light of frequency 
4
6.63
×10
14
 Hz is produced by source of 
power 2 mW . How many photons per second are emitted? 
Solution : ?? =
4
6.63
×10
14
 Hz 
?? =???? ??? =
?? ?? 
P=2×10
-3
 W 
1sec=2×10
-3
 J 
2×10
-3
=n
hc
?? ? 2×10
-3
=n×6.63×10
-34
×
4
6.63
×10
14
 
n=
1
2
×10
17
=5×10
16
 
n=5×10
16
 
Momentum of a photon : 
Example
?? =
?? ?? E.P. =
h
?? =
6.63×10
-34
3×10
8
+
4×10
14
6.63
=
4
3
×10
-28
=
2×10
-3
5×10
16
×3×10
8
=
4×10
-4
3×10
24
=
4
3
×10
-28
 
T. Source is of power =W. Find force exerted on wall if 
?? ?
?
?
 
(a) all photons are absorbed. 
(b) If all photons are reflected 
(c) 60% absorbed and 40% reflested. 
(d) All reflected 
Solution : (a) Let ?? be wavelength of light and ?? photons per sec are emitted. 
I sec? W Joule 
W=N(
hc
?? ) ? 
h
?? =
W
Nc
 
?? =?? |??? |=|0-
?? h
?? |=
?? ?? 
(b) ?? =|
-Nh
?? -
Nh
?? |=
2 W
c
 
(c) ?? =|
-40
100
Nh
?? -
Nh
?? |=
Nh
?? |0.4+01|?F=1.4
W
c
 
OR 
F=
60
100
W
c
-
40
100
(
2 W
c
) [
W
c
=100% absorbed ;
2 W
c
=100% reflected ] 
=
0.6 W
c
+
0.8 W
c
=1.4
W
c
 
(d) ?? =|-2
Nh
?? cos 60
°
|=
W
c
 
 
Work Function of a substance (?? ) : 
Minimum energy required to exit an electron from surface of the substance is called 
work function of the substance. 
Photoelectric Effect : 
Light is incident on metals surface with work function (?? ) . 
A photon may or may not collide with on electron. 
Consider a situation in which a photon collide with N electron and gives 100% of its 
energy to that electron now 
 
 
this energized electron may or may not collide with the lattice. 
Therefore its kinetic energy will be between 0 to (h?? -?? ) . 
0=KE=h?? -?? 
KE
max
=h?? -?? 
This equation is Einstein equation of photoelectric effect. 
When light hits a surface and makes electrons come off it, it's called the photoelectric 
effect. The electrons that get ejected during this are called photoelectrons. 
The minimum frequency for electron just comes out is called Threshold frequency (?? 0
) 
h?? 0
=?? 
The photoelectric effect takes place for V=?? 0
 
KE
max 
=h?? -?? =h?? -h?? 0
 
KE
max 
=n(?? -?? 0
) 
 
Maximum wavelength for which e
-
just comes out is called threshold wavelength (?? 0
)