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,
• In section 2.5, we saw the 'wave nature of the electromagnetic radiations' and the 'electromagnetic spectrum'
• In the previous section 2.6, we saw how 'particle nature' came to be known to the scientific community.
• In this section, we will see how Albert Einstein proved the 'particle nature'
1. Consider the experiment shown in fig.2.19(a) below:
• A metal surface is connected to the negative terminal of the battery
• A detector is connected to the positive terminal of the battery
• Both the metal surface and the detector are enclosed in a glass tube
• An ammeter is connected in the circuit so that, any flow of current can be detected
• In fig.a, the circuit is an open circuit. No current is flowing
2. Now consider fig.b
• A beam of light falls on the metal surface
• As soon as the light falls on the metal surface, electrons are ejected from it
• These electrons move towards the detector because it is positively charged
• Thus a flow of electrons occur and the circuit becomes complete
• The movement of the needle of the ammeter is the evidence of ‘electron flow’
■ We can write:
Electrons are ejected when metals like potassium, rubidium, caesium etc., are exposed to a beam of light. This phenomenon is called photoelectric effect
3. This experiment was first performed in 1887 by the German scientist Heinrich Hertz
Following are the three results obtained by him:
(i) Consider the instant at which the beam strikes the metal surface
• The needle of the ammeter deflects at the same instant
• That means, as soon as the beam strikes the metal surface, electrons are emitted
(ii) The number of electrons ejected is proportional to the brightness of the light
• That means:
♦ If the beam of light is bright, more electrons are ejected
♦ If the beam of light is dim, less electrons are ejected
(iii) consider the beam of light which falls on the metal surface
• It may be a light of low frequency like red or orange
• It may be a light of high frequency like blue or violet
■ But each metal has a unique frequency value
• The light falling on it must have a frequency equal to or greater than that unique value. Otherwise, electrons will not be ejected
■ We can write:
Each metal has a characteristic minimum frequency below which photoelectric effect is not observed. This frequency is called threshold frequency. It’s symbol is $\mathbf\small{\nu_0}$
• Imagine that a light of frequency $\mathbf\small{\nu_1}$ falls on a metal surface
♦ Let $\mathbf\small{\nu_1}$ be greater than the $\mathbf\small{\nu_0}$ of that metal
♦ Then the electrons will be surely ejected and those ejected electrons will possess some kinetic energy K1
• Imagine that another light of frequency $\mathbf\small{\nu_2}$ falls on that metal surface
♦ Let $\mathbf\small{\nu_2}$ be greater than $\mathbf\small{\nu_1}$
♦ Then also, the electrons will be surely ejected and those ejected electrons will possess some kinetic energy K2, where K2 is greater than K1
■ So we can write:
If the frequency of the light is increased, the kinetic energy (K) of the ejected electrons will also increase
4. The above three results in (3) could not be explained using the laws of classical physics
Let us see how classical physics fails in this case:
• According to the laws of classical physics, the energy content of a beam of light depends on the brightness of the light
• So, the following two theoretical results would be expected:
(i) If the beam of light is bright, greater number of electrons would be ejected
♦ If the beam of light is dim, lesser number of electrons would be ejected
(ii) If the beam of light is bright, the kinetic energy of the ejected electrons would be high
♦ If the beam of light is dim, the kinetic energy of the ejected electrons would be low
5. The theoretical result mentioned in [4(i)] agrees with the experimental results
♦ But the theoretical result mentioned in [4(ii)] does not agree with the experimental results
• Let us see an example. We can write it in two steps:
(i) Red light has a low frequency range (4.3 to 4.6 Hz)
• Let a very bright beam of red light fall on a piece of potassium metal
• Even if the potassium is exposed to this bright red light for hours, no electron will be ejected
(ii) Now consider yellow light. It has a higher frequency range (5.1 to 5.2 Hz)
• As soon as a dim beam of yellow light falls on the potassium metal, electrons are ejected
• Also, more electrons will be ejected if the brightness of that yellow light is increased
• But the increase in brightness does not increase the kinetic energy of the ejected electrons
6. So brightness is not the only criterion
The 'inability of even bright lights to initiate photoelectric effect' cannot be explained based on the laws of classical physics
• Similarly, the ‘unchanging kinetic energy even when brightness is increased’ cannot be explained using laws of classical physics
1. A beam of light contains packets of energy
• So, when a beam of light is directed towards a metal surface, we are shooting the metal with particles (packets)
2. When a packet strikes an electron, that electron will be knocked away from the atom
• That electron thus becomes free and is available to flow in the circuit
3. We have seen that, the energy contained in a packet depends on the frequency of the light
• So packets in the ‘low frequency lights (like red)’ will have only a small amount of energy
• Such packets cannot knock off electrons
4. But ‘packets in high frequency radiations’ can knock off electrons
• The energy ($\mathbf\small{h\nu}$) in such packets can be split into two parts:
Part 1: The first part will be used to knock off the electrons
• If $\mathbf\small{\nu_0}$ is the threshold frequency, this first part will be equal to $\mathbf\small{h\nu_0}$
Part 2: So the remaining energy after ‘knocking off’ is ($\mathbf\small{h\nu-h\nu_0}$)
• This remaining energy is the second part
• This remaining energy gets transformed into the kinetic energy (K) of the ‘knocked off electron’
• If me is the mass of the ‘knocked off electron’ and v the velocity, we can write: $\mathbf\small{K=\frac{1}{2}m_ev^2}$
5. Thus we get: $\mathbf\small{h\nu-h\nu_0=K}$
Rearranging this, we get:
Eq.2.2: $\mathbf\small{h\nu=h\nu_0+K}$
6. If the frequency of the incoming radiation is less than $\mathbf\small{\nu_0}$, there will not be any part 1 or part 2
• Because then, no electron will be knocked off
• We already saw that the minimum frequency required is $\mathbf\small{\nu_0}$, which is called the threshold frequency
• The ‘minimum energy’ corresponding to this ‘minimum required frequency’ is called 'work function'
■ So we can write:
The minimum energy required to knock off an electron is called work function. It’s symbol is W0
• So we get: W0 = $\mathbf\small{h\nu_0}$
7. The 'ejection of greater number of electrons' can be explained as follows:
(i) Consider a high intensity beam of light
('high intensity light' means 'bright light')
(ii) A bright light will contain a greater number of particles than dim light
(iii) So bright light will knock off a greater number of electrons
(i) We saw that radiations consist of ‘packets of energy’
• The name of this packet is photon
(ii) We saw that the ‘energy contained inside one packet’ is $\mathbf\small{h\nu}$
• The name of this ‘energy contained inside one packet’ is quantum
• The plural form of ‘quantum’ is quanta
■ We will write some examples:
• One photon of red light will contain ‘one quantum of energy’, which is equal to $\mathbf\small{h\nu_{Red}}$
• Two photons of red light will contain ‘two quanta of energy’, which is equal to $\mathbf\small{2h\nu_{Red}}$
• One photon of ultra violet radiation will contain ‘one quantum of energy’, which is equal to $\mathbf\small{h\nu_{UV}}$
• Two photons of ultra violet radiation will contain ‘two quanta of energy’, which is equal to $\mathbf\small{2h\nu_{UV}}$
(iii) The electrons ejected from the metal surface, as a result of the bombardment by photons are called photoelectrons
Solved examples 2.23 to 2.34
Solved examples 2.35 to 2.15
1. When the explanations for black body radiation and photoelectric effect were obtained, the 'particle nature' came to be known to the scientific community
2. But still there was confusion:
• Wave nature can explain diffraction and interference
♦ But wave nature cannot explain black body radiation and photoelectric effect
• Particle nature can explain black body radiation and photoelectric effect
♦ But particle nature cannot explain diffraction and interference
3. After much debate, scientists accepted the idea that, electromagnetic radiations possess both wave nature and particle nature
■ In other words:
Scientists accepted the idea that, electromagnetic radiations have dual behaviour
4. The behavior depends on the type of experiment
• If in an experiment, the radiation has to interact with matter, it will behave as a stream of particles
• If in an experiment, the radiation has to propagate only, it will behave as a wave
• In section 2.5, we saw the 'wave nature of the electromagnetic radiations' and the 'electromagnetic spectrum'
• In the previous section 2.6, we saw how 'particle nature' came to be known to the scientific community.
• In this section, we will see how Albert Einstein proved the 'particle nature'
1. Consider the experiment shown in fig.2.19(a) below:
 |
| Fig.2.19 |
• A detector is connected to the positive terminal of the battery
• Both the metal surface and the detector are enclosed in a glass tube
• An ammeter is connected in the circuit so that, any flow of current can be detected
• In fig.a, the circuit is an open circuit. No current is flowing
2. Now consider fig.b
• A beam of light falls on the metal surface
• As soon as the light falls on the metal surface, electrons are ejected from it
• These electrons move towards the detector because it is positively charged
• Thus a flow of electrons occur and the circuit becomes complete
• The movement of the needle of the ammeter is the evidence of ‘electron flow’
■ We can write:
Electrons are ejected when metals like potassium, rubidium, caesium etc., are exposed to a beam of light. This phenomenon is called photoelectric effect
3. This experiment was first performed in 1887 by the German scientist Heinrich Hertz
Following are the three results obtained by him:
(i) Consider the instant at which the beam strikes the metal surface
• The needle of the ammeter deflects at the same instant
• That means, as soon as the beam strikes the metal surface, electrons are emitted
(ii) The number of electrons ejected is proportional to the brightness of the light
• That means:
♦ If the beam of light is bright, more electrons are ejected
♦ If the beam of light is dim, less electrons are ejected
(iii) consider the beam of light which falls on the metal surface
• It may be a light of low frequency like red or orange
• It may be a light of high frequency like blue or violet
■ But each metal has a unique frequency value
• The light falling on it must have a frequency equal to or greater than that unique value. Otherwise, electrons will not be ejected
■ We can write:
Each metal has a characteristic minimum frequency below which photoelectric effect is not observed. This frequency is called threshold frequency. It’s symbol is $\mathbf\small{\nu_0}$
• Imagine that a light of frequency $\mathbf\small{\nu_1}$ falls on a metal surface
♦ Let $\mathbf\small{\nu_1}$ be greater than the $\mathbf\small{\nu_0}$ of that metal
♦ Then the electrons will be surely ejected and those ejected electrons will possess some kinetic energy K1
• Imagine that another light of frequency $\mathbf\small{\nu_2}$ falls on that metal surface
♦ Let $\mathbf\small{\nu_2}$ be greater than $\mathbf\small{\nu_1}$
♦ Then also, the electrons will be surely ejected and those ejected electrons will possess some kinetic energy K2, where K2 is greater than K1
■ So we can write:
If the frequency of the light is increased, the kinetic energy (K) of the ejected electrons will also increase
4. The above three results in (3) could not be explained using the laws of classical physics
Let us see how classical physics fails in this case:
• According to the laws of classical physics, the energy content of a beam of light depends on the brightness of the light
• So, the following two theoretical results would be expected:
(i) If the beam of light is bright, greater number of electrons would be ejected
♦ If the beam of light is dim, lesser number of electrons would be ejected
(ii) If the beam of light is bright, the kinetic energy of the ejected electrons would be high
♦ If the beam of light is dim, the kinetic energy of the ejected electrons would be low
5. The theoretical result mentioned in [4(i)] agrees with the experimental results
♦ But the theoretical result mentioned in [4(ii)] does not agree with the experimental results
• Let us see an example. We can write it in two steps:
(i) Red light has a low frequency range (4.3 to 4.6 Hz)
• Let a very bright beam of red light fall on a piece of potassium metal
• Even if the potassium is exposed to this bright red light for hours, no electron will be ejected
(ii) Now consider yellow light. It has a higher frequency range (5.1 to 5.2 Hz)
• As soon as a dim beam of yellow light falls on the potassium metal, electrons are ejected
• Also, more electrons will be ejected if the brightness of that yellow light is increased
• But the increase in brightness does not increase the kinetic energy of the ejected electrons
6. So brightness is not the only criterion
The 'inability of even bright lights to initiate photoelectric effect' cannot be explained based on the laws of classical physics
• Similarly, the ‘unchanging kinetic energy even when brightness is increased’ cannot be explained using laws of classical physics
The German scientist Albert Einstein was able to explain photoelectric effect in 1905. His explanation was based on Planck’s quantum theory that we saw in the previous section. The explanation can be detailed in seven steps as follows:
• So, when a beam of light is directed towards a metal surface, we are shooting the metal with particles (packets)
2. When a packet strikes an electron, that electron will be knocked away from the atom
• That electron thus becomes free and is available to flow in the circuit
3. We have seen that, the energy contained in a packet depends on the frequency of the light
• So packets in the ‘low frequency lights (like red)’ will have only a small amount of energy
• Such packets cannot knock off electrons
4. But ‘packets in high frequency radiations’ can knock off electrons
• The energy ($\mathbf\small{h\nu}$) in such packets can be split into two parts:
Part 1: The first part will be used to knock off the electrons
• If $\mathbf\small{\nu_0}$ is the threshold frequency, this first part will be equal to $\mathbf\small{h\nu_0}$
Part 2: So the remaining energy after ‘knocking off’ is ($\mathbf\small{h\nu-h\nu_0}$)
• This remaining energy is the second part
• This remaining energy gets transformed into the kinetic energy (K) of the ‘knocked off electron’
• If me is the mass of the ‘knocked off electron’ and v the velocity, we can write: $\mathbf\small{K=\frac{1}{2}m_ev^2}$
5. Thus we get: $\mathbf\small{h\nu-h\nu_0=K}$
Rearranging this, we get:
Eq.2.2: $\mathbf\small{h\nu=h\nu_0+K}$
6. If the frequency of the incoming radiation is less than $\mathbf\small{\nu_0}$, there will not be any part 1 or part 2
• Because then, no electron will be knocked off
• We already saw that the minimum frequency required is $\mathbf\small{\nu_0}$, which is called the threshold frequency
• The ‘minimum energy’ corresponding to this ‘minimum required frequency’ is called 'work function'
■ So we can write:
The minimum energy required to knock off an electron is called work function. It’s symbol is W0
• So we get: W0 = $\mathbf\small{h\nu_0}$
7. The 'ejection of greater number of electrons' can be explained as follows:
(i) Consider a high intensity beam of light
('high intensity light' means 'bright light')
(ii) A bright light will contain a greater number of particles than dim light
(iii) So bright light will knock off a greater number of electrons
Now we will see some official names:
• The name of this packet is photon
(ii) We saw that the ‘energy contained inside one packet’ is $\mathbf\small{h\nu}$
• The name of this ‘energy contained inside one packet’ is quantum
• The plural form of ‘quantum’ is quanta
■ We will write some examples:
• One photon of red light will contain ‘one quantum of energy’, which is equal to $\mathbf\small{h\nu_{Red}}$
• Two photons of red light will contain ‘two quanta of energy’, which is equal to $\mathbf\small{2h\nu_{Red}}$
• One photon of ultra violet radiation will contain ‘one quantum of energy’, which is equal to $\mathbf\small{h\nu_{UV}}$
• Two photons of ultra violet radiation will contain ‘two quanta of energy’, which is equal to $\mathbf\small{2h\nu_{UV}}$
(iii) The electrons ejected from the metal surface, as a result of the bombardment by photons are called photoelectrons
The links given below can be used to see some solved examples presented in pdf format
Solved examples 2.35 to 2.15
Dual behaviour of electromagnetic radiations
2. But still there was confusion:
• Wave nature can explain diffraction and interference
♦ But wave nature cannot explain black body radiation and photoelectric effect
• Particle nature can explain black body radiation and photoelectric effect
♦ But particle nature cannot explain diffraction and interference
3. After much debate, scientists accepted the idea that, electromagnetic radiations possess both wave nature and particle nature
■ In other words:
Scientists accepted the idea that, electromagnetic radiations have dual behaviour
4. The behavior depends on the type of experiment
• If in an experiment, the radiation has to interact with matter, it will behave as a stream of particles
• If in an experiment, the radiation has to propagate only, it will behave as a wave
We have completed a discussion on the basics of photoelectric effect. In the next section, we will see atomic spectra
No comments:
Post a Comment