We want to learn about the Bohr model of atom. For that, we must first learn about 'wave nature' and 'particle nature' of electromagnetic radiations. In this regard, we saw the 'wave nature of the electromagnetic radiations' and the 'electromagnetic spectrum' in the previous section. In this section, we will see the 'particle nature'
1. Consider the phenomenon of diffraction
• Diffraction is the bending of electromagnetic radiations around obstacles
• Scientists were able to give an explanation for this phenomenon
• The ‘wave nature of electromagnetic radiations’ was used to give the explanation
2. Similar is the case of the phenomenon of interference
• It deals with the interaction between two or more radiations
• Scientists were able to give an explanation for this phenomenon also
• Here also, they used the ‘wave nature of electromagnetic radiations’ to give the explanation
• We will learn more details about diffraction and interference in physics classes
3. Now consider the phenomenon of black body radiation
• Initially scientists were not able to give a satisfactory explanation for this phenomenon
• This is because, radiations were considered to be propagating in the form of waves only
• It is not possible to give an explanation for this phenomenon using ‘wave nature’
• ‘Particle nature’ of radiations was not known at that time
• In fact, the investigations into the phenomenon of black body radiations paved the way for the discovery of the 'particle nature'
4. So let us learn some basic details about black body radiation
• A black body is a body which absorbs all radiations that fall on it
(i) Consider a black body kept inside a dark room
• Apply a beam of green light on it
(ii) All those green light waves will be absorbed
• No wave will get reflected
(iii) We will be able to see any body only if ‘some light coming from that body’ falls in our eyes
• But here, no green wave is reflected. Also there is no other light
• So, when we apply green light, the body will appear black
(iv) Similar is the case with all other colors. We can apply red, blue etc.,
• The body will still be black
(v) If we apply white light, all the seven colors will be absorbed
• So instead of a dark room, if we place the black body in broad sunlight, still, it will appear black
(vi) It is not just the case of visible light. The black body will absorb all radiations in the electromagnetic spectrum
(vii) Also, the black body will not transmit any radiations. So no incident radiation will ever come out of a black body
5. True black bodies do not exist in nature. Carbon black can be considered as a close approximation
• For conducting experiments (on black body absorbsion) in the lab, scientists set up an apparatus as follows:
(i) A block with a cavity is taken
(ii) The only connection between the ‘interior of the cavity’ and the ‘outside environment’ is through a very small hole in the block
(iii) The inside surface of the cavity is coated with a layer of charcoal
(iv) The set up is complete
• Now apply a radiation through the small hole
(v) The radiation will enter the cavity through the hole
• Once it enters the cavity, it will suffer multiple reflections and will eventually get absorbed
• This is shown in fig.2.14 below:
(vi) Also, since the hole is very small, none of the reflections will have a chance to come out
• So this block acts as a black body
6. We saw that, the black body absorbs all radiations. So it will appear black
• Imagine that, a black body kept inside a dark room has become red in color
• Is there such a possibility of 'a black body becoming red'?
Let us check:
If we see a black body in red color, the following four points can be written:
(i) A black body never reflects any light or any radiations
(ii) So this red light is not any reflected light
(iii) This red light is coming from inside the black body
(iv) In other words, this black body is emitting red light on it’s own
7. But how is it possible for a black body to emit red light on it's own ?
The answer can be written in five steps:
(i) Heat the black body so that it’s temperature increases
• It’s atoms will begin to oscillate
• So the electrons in the atoms will also oscillate
(ii) When the electrons oscillate, they emit electromagnetic radiations
(iii) If the wavelength of this emitted radiation is in the range of 750 – 610 nm, we will see red light
• If the wavelength of this emitted radiation is in the range of 610 – 590 nm, we will see orange light
so on . . .
(iv) Thus we can say:
• Heating is an effective method to make a black body to emit light.
(v) In fact, if the surrounding environmental conditions are favorable, a black body can emit invisible radiations also
8. So it is clear that, 'heating' can cause bodies to emit light
• We see several examples around us. Let us see one example. We will write it in steps:
(i) Consider an iron rod being heated in a furnace
• A furnace is capable of creating very high temperatures
(ii) Initially, the iron rod has a black or gray color
• But as the temperature increases, we begin to observe some changes. We will write it in stages:
Stage 1:
• When the temperature is low, the iron is a little hot. But not very hot
• The iron still has the original black/gray color
• Even if the color has not changed, the emission of radiations have begun
• The radiations at this stage are in the infrared range
• We know that infrared radiations are in fact heat radiations
• So even if we do not touch the rod, we will feel the heat. All we need to do is bring our hands closer to the rod
Stage 2:
• The temperature is increased further
• The iron appears in a dull red color
• This is visible evidence that, the emission of radiations have begun
• The emitted radiations at this stage are in the red region of the visible spectrum
Stage 3:
• The temperature is increased further
• The iron appears in a orange color
• The emitted radiations at this stage are in the orange region of the visible spectrum
Stage 4:
• The temperature is increased further
• The iron appears in a yellow color
• The emitted radiations at this stage are in the yellow region of the visible spectrum
_ _ _ _ _
_ _ _ _ _
• This goes on and we reach the
Final stage:
• The temperature is increased to a very high value
• The iron appears in a bluish white color
• The emitted radiations at this stage are in the left extreme region of the visible spectrum
9. Based on (8), we can say, black bodies are indeed able to emit light
• 'Emit' is the opposite of 'absorb'
• Earlier we saw that, the black body can absorb all radiations in the electromagnetic spectrum
■ The converse (opposite) is also true:
A black body can emit all radiations in the electromagnetic spectrum
10. For conducting experiments (on black body emission) in the lab, scientists set up the apparatus as follows:
(i) Consider an oven
(ii) It has a 'heating element'
• The 'heating element' is placed in the interior of the oven
(iii) When the heating element becomes hot, the interior of the oven also becomes hot
• The exterior of the oven remains at room temperature
(iv) Scientists put a very small hole on the side of the oven
• If we look through the hole (after taking all safety precautions to protect the eyes and other parts of our body), we can see the heating element changing it’s color from red to bluish white progressively, with increase in temperature
11. Scientists measure the intensity of the radiations coming out through the hole
• A graph is plotted with wavelength along the x axis and intensity along the y axis
■ Intensity is defined as the ‘amount of energy passing through unit area’
• For our present discussion, intensity can be considered simply as ‘amount of energy’
12. What is the purpose of a intensity vs wavelength graph ?
The answer can be written in 7 steps:
(i) Consider the graph shown in fig.2.15(a) below:
• It is plotted using the 'observations made when the temperature of the black body is 6000 K'
(Note that, high temperatures like 6000 K cannot be obtained in the lab. These graphs are plotted using the data obtained from the observations of distant stars)
(ii) Let us see what information we get from this graph
• As an example, in fig.2.15 (b), consider the vertical white dotted line drawn in between 500 and 1000 nm
• Let this dotted line intersect the x axis at $\mathbf\small{\lambda=\lambda_1}$
(iii) Then we can write:
The energy contributed by the radiation of wavelength $\mathbf\small{\lambda_1}$
= height of that dotted line
(iv) Another example: Consider the vertical white dotted line drawn in between 1000 and 1500 nm.
Let this dotted line intersect the x axis at $\mathbf\small{\lambda=\lambda_2}$.
(v) Then we can write:
The energy contributed by the radiation of wavelength $\mathbf\small{\lambda_2}$
= height of that dotted line
(vi) We see that:
• Radiations with lower wavelengths make lesser contributions.
• Radiations with higher wavelengths also make lesser contributions.
• Radiations with intermediate wavelengths make greater contributions.
(vii) The above graph shows the energy values when the temperature is 6000 K
• When the temperature is 7000 K, we get a similar graph (shown in blue color in fig,2.16.a), but with greater height
• The greater height indicates that, more energy is radiated out. This is natural because:
If the body is at a higher temperature, it means that there is more ‘energy content’. The body can radiate out more energy
13. Scientists of the latter half of the nineteenth century tried to find the relations between
the three items: (i) radiated energy (ii) temperature (iii) wavelength
• Based on their findings, a 'theoretical graph' was plotted
• This graph is shown as a cyan dashed curve in fig.2.16 (b)
14. On the right side, this graph is in agreement with the experimental results
• But when we come towards the left, this graph continues to move upwards
(Note that, the actual graph reaches a peak value and then drops)
• According to the theoretical cyan dashed curve, we will get (when temperature increases) radiations of very small wavelengths
15. Radiations of very small wavelengths means:
Radiations of very high frequencies (like X-rays and gamma rays)
• But we never get 'significant quantities of those high frequency waves' from a hot body even if the temperature is very high
• So the theoretical cyan dashed curve is wrong
16. Thus scientists faced the problems:
• Why is the cyan dashed curve wrong?
• What is causing the red and blue curves to peak and then fall down?
17. The explanation was given by the German scientist Max Planck
• The cyan dashed curve is plotted based on the assumption that, radiations propagate in the form of waves
■ Planck proposed that, radiations propagate as 'packets of energy'
• Based on his studies, Planck plotted his own theoretical curve. It is shown as the green dashed curve in fig.2.16(b)
• We see that, this theoretical curve is in agreement with the curves obtained experimentally
(We will see the 'actual equations used for the theoretical curves' in higher classes)
18. Let us write a summary. It can be written in two steps:
(i) The cyan dashed curve in fig.2.16(b) is a theoretical curve
• It is based on the assumption that, electromagnetic radiations propagate in the form of waves
• This cyan dashed curve does not agree with the curves plotted based on experimental data
(ii) The green dashed curve in fig.2.16(b) is also a theoretical curve
• It is based on the assumption that, electromagnetic radiations propagate in the form of packets of energy
• This green dashed curve agrees with the curves plotted based on experimental data
19. So the scientific community had to accept the fact that, electromagnetic waves indeed possess particle nature also
So we will learn about those findings in some more detail:
■ Max Planck proposed that, electromagnetic radiations propagate in the form of 'packets of energy'. Let us elaborate on this:
1. Consider any radiation in the electromagnetic spectrum. Let the frequency of that radiation be 𝜈
2. That radiation will consist of 'packets of energy'
• All packets will contain the same ‘quantity of energy’
3. This ‘quantity of energy’ contained inside any one packet is called quantum
♦ Note that, ‘quantum’ is not the ‘name of the packet’
♦ It is the ‘quantity of energy’ contained inside the packet
• Quantum is given the symbol 'E'
4. This ‘quantum’ is proportional to the 𝜈 of that radiation
• Planck gave the following relation:
One quantum in any radiation = E = h𝜈
Where:
• h is called the Planck’s constant
♦ It has a value of 6.636 × 10-34
♦ It's unit is J s
♦ So we can write: h = 6.636 × 10-34 J s
• 𝜈 is the frequency of that radiation
5. We can write:
The quantity of energy contained inside one packet of a radiation (with frequency 𝜈)
= One quantum
= E = h𝜈
6. All packets in a particular radiation will be identical because, 𝜈 is a constant for a radiation
• But packets of a radiation will be different from the packets of another radiation
• For example a packet of red will be smaller than a packet of orange because, orange has a greater frequency
• We already mentioned that, the quantity of energy inside one packet is called quantum
• So we can write:
'One quantum of red' will be smaller than 'one quantum of orange'. This is shown in fig.2.17 below:
Solved example 2.22
Using the method of dimensional analysis, find the dimensions of 'h' in the relation E = h𝜈. Also find the units of 'h'.
Solution:
• 'E' is energy. So we have: $\mathbf\small{[E]=[ML^2T^{-2}]}$
• [𝜈] = [T-1]
• Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[ML^2T^{-2}]=[h][T^{-1}]}$
$\mathbf\small{\Rightarrow[h]=\frac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]}$
So 'h' has the dimensions: $\mathbf\small{[ML^2T^{-1}]}$
• To find the units, we will keep [E] as such
Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[E]=[h][T^{-1}]}$
$\mathbf\small{\Rightarrow[h]=\frac{[E]}{[T^{-1}]}=[E][T]}$
• Units of energy is joules
• Units of time is second
• So units of 'h' is J s
8. Planck's quantum theory states that, energy can be emitted only in packets
• Imagine that, red light is coming out from a body
• The total energy that is emitted from the body will be equal to [nh(𝜈Red)]
• Where
♦ 'n' is the number of packets emitted from the body
♦ 'n' should be an integer like 1, 2, 3, etc
• Remember that, 'n' should be an integer. It should not be given values like 1.2, 3.4, 72⁄3 etc., This is because, we cannot subdivide a packet
9. Absorption of energy also takes place in packets
• So in general, we can write:
Quantity of energy emitted or absorbed = nh𝜈
• Where:
♦ 'n' is the number of packets emitted or absorbed
♦ h𝜈 is the quantity of energy contained inside one packet
10. Let us plot two more graphs. They correspond to 5000 and 4000 K
• The modified graph is shown in fig.2.18(a) below:
• We see that, as the temperature decreases, the curves become flatter
• This is because, as the temperature decreases, the 'energy radiated out' decreases
11. Consider the peak of each of the four curves
• As the temperature increases, the peaks shift towards the left
• We can write:
As the temperature increases, the peaks shift towards higher frequencies
• That means:
As the temperature increases, ‘radiations of higher frequencies’ will be emitted more
• That is why we see a bluish tint when the iron rod becomes very hot
12. In the previous section, we plotted the visible spectrum on the '$\mathbf\small{\lambda}$ axis' (Fig. 2.10)
• Here also, in fig.2.18(a), we have a '$\mathbf\small{\lambda}$ axis'. So let us plot it again. The result is shown in fig.b
We see that, when the temperature increases, the peak indeed tends towards the blue side
■ Consider a black body which is at a high temperature
• It begins to give off it’s energy in the form of radiations
• Since the temperature is high, a large quantity of energy is available to be radiated out
• The body has to face the question:
How to give off the 'large energy’?
• The body has two options
(i) Give off a large number of small packets
(ii) Give off a small number of large packets
• Obviously, option (ii) is easier
• So we get large packets from the body
• That means, we get radiations of higher frequencies
• In other words, we get radiations which belong to the blue side
Planck's Quantum theory
1. Consider the phenomenon of diffraction
• Diffraction is the bending of electromagnetic radiations around obstacles
• Scientists were able to give an explanation for this phenomenon
• The ‘wave nature of electromagnetic radiations’ was used to give the explanation
2. Similar is the case of the phenomenon of interference
• It deals with the interaction between two or more radiations
• Scientists were able to give an explanation for this phenomenon also
• Here also, they used the ‘wave nature of electromagnetic radiations’ to give the explanation
• We will learn more details about diffraction and interference in physics classes
3. Now consider the phenomenon of black body radiation
• Initially scientists were not able to give a satisfactory explanation for this phenomenon
• This is because, radiations were considered to be propagating in the form of waves only
• It is not possible to give an explanation for this phenomenon using ‘wave nature’
• ‘Particle nature’ of radiations was not known at that time
• In fact, the investigations into the phenomenon of black body radiations paved the way for the discovery of the 'particle nature'
4. So let us learn some basic details about black body radiation
• A black body is a body which absorbs all radiations that fall on it
(i) Consider a black body kept inside a dark room
• Apply a beam of green light on it
(ii) All those green light waves will be absorbed
• No wave will get reflected
(iii) We will be able to see any body only if ‘some light coming from that body’ falls in our eyes
• But here, no green wave is reflected. Also there is no other light
• So, when we apply green light, the body will appear black
(iv) Similar is the case with all other colors. We can apply red, blue etc.,
• The body will still be black
(v) If we apply white light, all the seven colors will be absorbed
• So instead of a dark room, if we place the black body in broad sunlight, still, it will appear black
(vi) It is not just the case of visible light. The black body will absorb all radiations in the electromagnetic spectrum
(vii) Also, the black body will not transmit any radiations. So no incident radiation will ever come out of a black body
5. True black bodies do not exist in nature. Carbon black can be considered as a close approximation
• For conducting experiments (on black body absorbsion) in the lab, scientists set up an apparatus as follows:
(i) A block with a cavity is taken
(ii) The only connection between the ‘interior of the cavity’ and the ‘outside environment’ is through a very small hole in the block
(iii) The inside surface of the cavity is coated with a layer of charcoal
(iv) The set up is complete
• Now apply a radiation through the small hole
(v) The radiation will enter the cavity through the hole
• Once it enters the cavity, it will suffer multiple reflections and will eventually get absorbed
• This is shown in fig.2.14 below:
 |
| Fig.2.14 |
• So this block acts as a black body
6. We saw that, the black body absorbs all radiations. So it will appear black
• Imagine that, a black body kept inside a dark room has become red in color
• Is there such a possibility of 'a black body becoming red'?
Let us check:
If we see a black body in red color, the following four points can be written:
(i) A black body never reflects any light or any radiations
(ii) So this red light is not any reflected light
(iii) This red light is coming from inside the black body
(iv) In other words, this black body is emitting red light on it’s own
7. But how is it possible for a black body to emit red light on it's own ?
The answer can be written in five steps:
(i) Heat the black body so that it’s temperature increases
• It’s atoms will begin to oscillate
• So the electrons in the atoms will also oscillate
(ii) When the electrons oscillate, they emit electromagnetic radiations
(iii) If the wavelength of this emitted radiation is in the range of 750 – 610 nm, we will see red light
• If the wavelength of this emitted radiation is in the range of 610 – 590 nm, we will see orange light
so on . . .
(iv) Thus we can say:
• Heating is an effective method to make a black body to emit light.
(v) In fact, if the surrounding environmental conditions are favorable, a black body can emit invisible radiations also
8. So it is clear that, 'heating' can cause bodies to emit light
• We see several examples around us. Let us see one example. We will write it in steps:
(i) Consider an iron rod being heated in a furnace
• A furnace is capable of creating very high temperatures
(ii) Initially, the iron rod has a black or gray color
• But as the temperature increases, we begin to observe some changes. We will write it in stages:
Stage 1:
• When the temperature is low, the iron is a little hot. But not very hot
• The iron still has the original black/gray color
• Even if the color has not changed, the emission of radiations have begun
• The radiations at this stage are in the infrared range
• We know that infrared radiations are in fact heat radiations
• So even if we do not touch the rod, we will feel the heat. All we need to do is bring our hands closer to the rod
Stage 2:
• The temperature is increased further
• The iron appears in a dull red color
• This is visible evidence that, the emission of radiations have begun
• The emitted radiations at this stage are in the red region of the visible spectrum
Stage 3:
• The temperature is increased further
• The iron appears in a orange color
• The emitted radiations at this stage are in the orange region of the visible spectrum
Stage 4:
• The temperature is increased further
• The iron appears in a yellow color
• The emitted radiations at this stage are in the yellow region of the visible spectrum
_ _ _ _ _
_ _ _ _ _
• This goes on and we reach the
Final stage:
• The temperature is increased to a very high value
• The iron appears in a bluish white color
• The emitted radiations at this stage are in the left extreme region of the visible spectrum
9. Based on (8), we can say, black bodies are indeed able to emit light
• 'Emit' is the opposite of 'absorb'
• Earlier we saw that, the black body can absorb all radiations in the electromagnetic spectrum
■ The converse (opposite) is also true:
A black body can emit all radiations in the electromagnetic spectrum
10. For conducting experiments (on black body emission) in the lab, scientists set up the apparatus as follows:
(i) Consider an oven
(ii) It has a 'heating element'
• The 'heating element' is placed in the interior of the oven
(iii) When the heating element becomes hot, the interior of the oven also becomes hot
• The exterior of the oven remains at room temperature
(iv) Scientists put a very small hole on the side of the oven
• If we look through the hole (after taking all safety precautions to protect the eyes and other parts of our body), we can see the heating element changing it’s color from red to bluish white progressively, with increase in temperature
11. Scientists measure the intensity of the radiations coming out through the hole
• A graph is plotted with wavelength along the x axis and intensity along the y axis
■ Intensity is defined as the ‘amount of energy passing through unit area’
• For our present discussion, intensity can be considered simply as ‘amount of energy’
12. What is the purpose of a intensity vs wavelength graph ?
The answer can be written in 7 steps:
(i) Consider the graph shown in fig.2.15(a) below:
 |
| Fig.2.15 |
(Note that, high temperatures like 6000 K cannot be obtained in the lab. These graphs are plotted using the data obtained from the observations of distant stars)
(ii) Let us see what information we get from this graph
• As an example, in fig.2.15 (b), consider the vertical white dotted line drawn in between 500 and 1000 nm
• Let this dotted line intersect the x axis at $\mathbf\small{\lambda=\lambda_1}$
(iii) Then we can write:
The energy contributed by the radiation of wavelength $\mathbf\small{\lambda_1}$
= height of that dotted line
(iv) Another example: Consider the vertical white dotted line drawn in between 1000 and 1500 nm.
Let this dotted line intersect the x axis at $\mathbf\small{\lambda=\lambda_2}$.
(v) Then we can write:
The energy contributed by the radiation of wavelength $\mathbf\small{\lambda_2}$
= height of that dotted line
(vi) We see that:
• Radiations with lower wavelengths make lesser contributions.
• Radiations with higher wavelengths also make lesser contributions.
• Radiations with intermediate wavelengths make greater contributions.
(vii) The above graph shows the energy values when the temperature is 6000 K
• When the temperature is 7000 K, we get a similar graph (shown in blue color in fig,2.16.a), but with greater height
• The greater height indicates that, more energy is radiated out. This is natural because:
If the body is at a higher temperature, it means that there is more ‘energy content’. The body can radiate out more energy
 |
| Fig.2.16 |
the three items: (i) radiated energy (ii) temperature (iii) wavelength
• Based on their findings, a 'theoretical graph' was plotted
• This graph is shown as a cyan dashed curve in fig.2.16 (b)
14. On the right side, this graph is in agreement with the experimental results
• But when we come towards the left, this graph continues to move upwards
(Note that, the actual graph reaches a peak value and then drops)
• According to the theoretical cyan dashed curve, we will get (when temperature increases) radiations of very small wavelengths
15. Radiations of very small wavelengths means:
Radiations of very high frequencies (like X-rays and gamma rays)
• But we never get 'significant quantities of those high frequency waves' from a hot body even if the temperature is very high
• So the theoretical cyan dashed curve is wrong
16. Thus scientists faced the problems:
• Why is the cyan dashed curve wrong?
• What is causing the red and blue curves to peak and then fall down?
17. The explanation was given by the German scientist Max Planck
• The cyan dashed curve is plotted based on the assumption that, radiations propagate in the form of waves
■ Planck proposed that, radiations propagate as 'packets of energy'
• Based on his studies, Planck plotted his own theoretical curve. It is shown as the green dashed curve in fig.2.16(b)
• We see that, this theoretical curve is in agreement with the curves obtained experimentally
(We will see the 'actual equations used for the theoretical curves' in higher classes)
18. Let us write a summary. It can be written in two steps:
(i) The cyan dashed curve in fig.2.16(b) is a theoretical curve
• It is based on the assumption that, electromagnetic radiations propagate in the form of waves
• This cyan dashed curve does not agree with the curves plotted based on experimental data
(ii) The green dashed curve in fig.2.16(b) is also a theoretical curve
• It is based on the assumption that, electromagnetic radiations propagate in the form of packets of energy
• This green dashed curve agrees with the curves plotted based on experimental data
19. So the scientific community had to accept the fact that, electromagnetic waves indeed possess particle nature also
The findings made by Max Planck played a major role in the development of modern physics
■ Max Planck proposed that, electromagnetic radiations propagate in the form of 'packets of energy'. Let us elaborate on this:
1. Consider any radiation in the electromagnetic spectrum. Let the frequency of that radiation be 𝜈
2. That radiation will consist of 'packets of energy'
• All packets will contain the same ‘quantity of energy’
3. This ‘quantity of energy’ contained inside any one packet is called quantum
♦ Note that, ‘quantum’ is not the ‘name of the packet’
♦ It is the ‘quantity of energy’ contained inside the packet
• Quantum is given the symbol 'E'
4. This ‘quantum’ is proportional to the 𝜈 of that radiation
• Planck gave the following relation:
One quantum in any radiation = E = h𝜈
Where:
• h is called the Planck’s constant
♦ It has a value of 6.636 × 10-34
♦ It's unit is J s
♦ So we can write: h = 6.636 × 10-34 J s
• 𝜈 is the frequency of that radiation
5. We can write:
The quantity of energy contained inside one packet of a radiation (with frequency 𝜈)
= One quantum
= E = h𝜈
6. All packets in a particular radiation will be identical because, 𝜈 is a constant for a radiation
• But packets of a radiation will be different from the packets of another radiation
• For example a packet of red will be smaller than a packet of orange because, orange has a greater frequency
• We already mentioned that, the quantity of energy inside one packet is called quantum
• So we can write:
'One quantum of red' will be smaller than 'one quantum of orange'. This is shown in fig.2.17 below:
 |
| Fig.2.17 |
Similarly, 'one quantum of radio waves' will be smaller than 'one quantum of red'
7. In (4) above, we wrote the units of 'h' as J s. Let us prove it using dimensional analysis. We will do it as a solved example
7. In (4) above, we wrote the units of 'h' as J s. Let us prove it using dimensional analysis. We will do it as a solved example
Solved example 2.22
Using the method of dimensional analysis, find the dimensions of 'h' in the relation E = h𝜈. Also find the units of 'h'.
Solution:
• 'E' is energy. So we have: $\mathbf\small{[E]=[ML^2T^{-2}]}$
• [𝜈] = [T-1]
• Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[ML^2T^{-2}]=[h][T^{-1}]}$
$\mathbf\small{\Rightarrow[h]=\frac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]}$
So 'h' has the dimensions: $\mathbf\small{[ML^2T^{-1}]}$
• To find the units, we will keep [E] as such
Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[E]=[h][T^{-1}]}$
$\mathbf\small{\Rightarrow[h]=\frac{[E]}{[T^{-1}]}=[E][T]}$
• Units of energy is joules
• Units of time is second
• So units of 'h' is J s
8. Planck's quantum theory states that, energy can be emitted only in packets
• Imagine that, red light is coming out from a body
• The total energy that is emitted from the body will be equal to [nh(𝜈Red)]
• Where
♦ 'n' is the number of packets emitted from the body
♦ 'n' should be an integer like 1, 2, 3, etc
• Remember that, 'n' should be an integer. It should not be given values like 1.2, 3.4, 72⁄3 etc., This is because, we cannot subdivide a packet
9. Absorption of energy also takes place in packets
• So in general, we can write:
Quantity of energy emitted or absorbed = nh𝜈
• Where:
♦ 'n' is the number of packets emitted or absorbed
♦ h𝜈 is the quantity of energy contained inside one packet
10. Let us plot two more graphs. They correspond to 5000 and 4000 K
• The modified graph is shown in fig.2.18(a) below:
 |
| Fig.2.18 |
• This is because, as the temperature decreases, the 'energy radiated out' decreases
11. Consider the peak of each of the four curves
• As the temperature increases, the peaks shift towards the left
• We can write:
As the temperature increases, the peaks shift towards higher frequencies
• That means:
As the temperature increases, ‘radiations of higher frequencies’ will be emitted more
• That is why we see a bluish tint when the iron rod becomes very hot
12. In the previous section, we plotted the visible spectrum on the '$\mathbf\small{\lambda}$ axis' (Fig. 2.10)
• Here also, in fig.2.18(a), we have a '$\mathbf\small{\lambda}$ axis'. So let us plot it again. The result is shown in fig.b
We see that, when the temperature increases, the peak indeed tends towards the blue side
■ Consider a black body which is at a high temperature
• It begins to give off it’s energy in the form of radiations
• Since the temperature is high, a large quantity of energy is available to be radiated out
• The body has to face the question:
How to give off the 'large energy’?
• The body has two options
(i) Give off a large number of small packets
(ii) Give off a small number of large packets
• Obviously, option (ii) is easier
• So we get large packets from the body
• That means, we get radiations of higher frequencies
• In other words, we get radiations which belong to the blue side
We have completed a discussion on the basics of black body radiation. This phenomenon and it’s explanation given by Max Planck, gave the scientific community, the first indications about the particle nature. Later Albert Einstein proved it beyond doubt that, radiations indeed have particle nature as well. This he did while explaining the photoelectric effect. We will see it in the next section
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