In the previous section, we completed a discussion on mole concept. In this chapter we will see structure of atom
• In ancient times, scientists and philosophers believed that atoms are the fundamental building blocks of matter
• Consider the following steps:
(i) If we cut an object into two equal parts, each will have a size half of the original. Let the original volume be k m3.
(ii) Then each of the two new pieces will have a volume of 0.5k m3
(iii) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.5k) = 0.25k m3
(iv) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.25k) = 0.125k m3
(v) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.125k) = 0.0625k m3
(vi) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.0625k) = 0.03125k m3
(vii) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.03125k) = 0.0.015625k m3
• We see that with each step, the particle becomes smaller and smaller
• If we continue like this, a stage will be reached where the particle obtained will become so small that it can no longer be cut
• Ancient scientists and philosophers called this tiny particle: atom
• The word atom is derived from the Greek word ‘a-tomio’ which means ‘non divisible’
• However, these earlier ideas related to 'non divisibility' were not proved experimentally
• What were those developments which enabled the scientists to question the ancient ideas ?
• We will discuss them in detail:
• It is called a cathode ray discharge tube
2. Two thin pieces of metal, one at each end, are sealed inside the tube
3. One of these metal pieces is connected to the negative terminal of the battery
• This piece is the cathode
4. The other piece is connected to the positive terminal of the battery
• This piece is the anode
5. We see that it is an open circuit. There is no way current can flow
6. But if a high voltage is applied, and the pressure inside the tube is low, current begins to flow
7. The current flow becomes possible because, a stream of particles move from the cathode to the anode
• This stream of particles is called cathode rays
■ Cathode rays are produced only when high voltages are applied
■ Also the pressure inside the tube must be sufficiently low
• Pressure can be reduced to the required level by using a vacuum pump
8. Visible evidence of the cathode rays can be obtained by making some simple modifications to the tube
• In fig.2.1 (b) below, the anode has a hole in it
• And the portion of the tube behind the anode is coated with zinc sulphide
9. When the high voltage is applied on this new device, a bright spot appears on the coating
• The reason for this bright spot can be written in 3 steps:
(i) Zinc sulphide is a phosphorescent material. It emits light when excited
(ii) So it is clear that, some form of energy is falling on the coating. But we see no physical contacts being made on the coating
(iii) The only explanation is this:
Some of the particles coming from the cathode, passes through the hole in the anode and hit the zinc sulphide coating
10. This bright spot is visible evidence that, some particles are indeed coming out from the cathode
• In ancient times, scientists and philosophers believed that atoms are the fundamental building blocks of matter
• Consider the following steps:
(i) If we cut an object into two equal parts, each will have a size half of the original. Let the original volume be k m3.
(ii) Then each of the two new pieces will have a volume of 0.5k m3
(iii) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.5k) = 0.25k m3
(iv) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.25k) = 0.125k m3
(v) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.125k) = 0.0625k m3
(vi) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.0625k) = 0.03125k m3
(vii) Take one of them and cut it into two equal parts. Each will have a volume of (0.5 × 0.03125k) = 0.0.015625k m3
• We see that with each step, the particle becomes smaller and smaller
• If we continue like this, a stage will be reached where the particle obtained will become so small that it can no longer be cut
• Ancient scientists and philosophers called this tiny particle: atom
• The word atom is derived from the Greek word ‘a-tomio’ which means ‘non divisible’
• However, these earlier ideas related to 'non divisibility' were not proved experimentally
• Developments in the nineteenth century enabled modern scientists to question the idea that atom is non divisible
• We will discuss them in detail:
Discovery of Electron
1. Fig.2.1 (a) below shows a glass tube. It is outlined in cyan color• It is called a cathode ray discharge tube
 |
| Fig.2.1 (a) |
3. One of these metal pieces is connected to the negative terminal of the battery
• This piece is the cathode
4. The other piece is connected to the positive terminal of the battery
• This piece is the anode
5. We see that it is an open circuit. There is no way current can flow
6. But if a high voltage is applied, and the pressure inside the tube is low, current begins to flow
7. The current flow becomes possible because, a stream of particles move from the cathode to the anode
• This stream of particles is called cathode rays
■ Cathode rays are produced only when high voltages are applied
■ Also the pressure inside the tube must be sufficiently low
• Pressure can be reduced to the required level by using a vacuum pump
8. Visible evidence of the cathode rays can be obtained by making some simple modifications to the tube
• In fig.2.1 (b) below, the anode has a hole in it
 |
| Fig.2.1 (b) |
9. When the high voltage is applied on this new device, a bright spot appears on the coating
• The reason for this bright spot can be written in 3 steps:
(i) Zinc sulphide is a phosphorescent material. It emits light when excited
(ii) So it is clear that, some form of energy is falling on the coating. But we see no physical contacts being made on the coating
(iii) The only explanation is this:
Some of the particles coming from the cathode, passes through the hole in the anode and hit the zinc sulphide coating
10. This bright spot is visible evidence that, some particles are indeed coming out from the cathode
The results of the above experiments can be summarized as follows:
1. The cathode rays start from cathode and move towards the anode
2. These are not visible rays
• Their presence can be detected using fluorescent or phosphorescent materials
3. This is just like creating pictures on a television screen
• Television picture tubes are cathode ray tubes coated with fluorescent or phosphorescent materials
• Different portions of the coating glow with different intensities when cathode rays fall on those portions with different intensities. This creates the required pictures on the screen
4. The cathode rays travel in a straight line if there are no electric or magnetic fields
5. In the presence of electric or magnetic fields, the cathode rays behave as if they are made up of negatively charged particles. We will see those details soon
• This helped scientists to conclude that, cathode rays consist of negatively charged particles
• They called those particles: electrons
6. The characteristics of the cathode rays do not depend upon the materials of the electrodes
• That is., even if electrodes made of different materials are used in the tube, the results of the experiment do not change
7. The characteristics of the cathode rays do not depend upon the gas present in the tube
• That is., even if any other gas is used in the tube, the results of the experiment do not change
8. The findings in (6) and (7) helped scientists to conclude that, nature of electrons do not depend on the type of materials
• In other words, electrons present in all materials have the same characteristics
1. A ray of electrons is shown in red color
• It starts from the cathode
• And passes through the hole in the anode
2. The ray flows along the x direction
(The 3 mutually perpendicular axes are shown at right side bottom of the diagram)
3. Two electrodes (shown in cyan color) are placed in the path of the ray
• When a voltage is applied across these electrodes, an electric field is created between them
• The electrodes are arranged in such a way that, the electric field is perpendicular to the electron flow
• From the fig., it is clear that, the electric field is in the z direction
4. In the fig., no voltage is applied and so there is no electric field
• Since there is no electric field, the ray passes without any deviation and hits the point B on the fluorescent screen
5. Let us see what happens when voltage is applied:
• In fig.2.2 (b) below, the upper electrode is connected to the positive terminal of the battery and the lower electrode to the negative terminal
• The ray then deviates upwards (in the z direction) and hits the screen at point A
6. In fig.2.2 (c) below, the voltage across the cyan electrodes is cut off. Instead, a magnetic field is applied
• The north pole of the magnet is shown in red color
• The south pole of the magnet is shown in blue color
• The poles are arranged in such a way that, the magnetic field is perpendicular to the electron flow
• From the fig., it is clear that, the magnetic field is in the y direction
• The ray then deviates downwards (in the z direction) and hits the screen at point C
7. The points A, B and C lie in the same vertical line
• Let us see the reason for 'C' being in the same vertical line as A and B:
(i) We have learnt about Fleming's right hand rule in our earlier classes (Details here)
• Consider the fig.2.2 (d) below. It shows the right arm in an inverted position
(ii) The forefinger points in the direction of the magnetic field (magenta arrow in the y direction)
• The middle finger points in the direction of the current (brown arrow in the x direction)
(iii) So the thump points in the direction of the 'deflection of the current carrying conductor' (cyan arrow in the downward z direction)
• Thus we see how point C comes vertically below A and B
8. Now we get back to the main discussion
• We saw these:
♦ When the electric field is applied, the ray deviates upwards
♦ When the magnetic field is applied, the ray deviates downwards
■ So let us see what happens when both the fields are applied simultaneously
• The electric field tries to pull the ray upwards
• But this pulling effect can be nullified by applying a magnetic field of suitable strength
• When this is done, the ray passes without any deviation and hits the point B
• This is shown in fig. e below:
9. This experiment was carried out by the British Physicist J.J. Thompson
• Three points were noted by him regarding the 'amount of deviation' from the normal path:
(i) The amount of deviation depends upon the magnitude of the charge of the particle
(ii) The amount of deviation depends upon the mass of the particle
(iii) The amount of deviation depends upon the strength of the fields
We will now see each of the 3 items in more detail
(i) Magnitude of the charge:
• We saw in fig. 2.2(a) that, when the electric field is applied, the ray deflects towards the positive electrode
• So we can conclude that, the particles are negatively charged
♦ If the magnitude of this negative charge is greater, the deflection will be greater
♦ If the magnitude of this negative charge is lesser, the deflection will be lesser
• So deflection (d) is directly proportional to the charge e
• We can write: d ∝ e
(The symbol '∝' represents 'is proportional to')
(ii) Mass of the particle
♦ If the mass is greater, deflection will be lesser
♦ If the mass is lesser, deflection will be greater
• So deflection (d) is inversely proportional to the mass m
♦ Where me is the mass of one electron
• We can write: $\mathbf\small{d\propto\frac{1}{m_e}}$
(iii) Strength of fields
♦ If the strength of the electric field is greater, the deflection (upward) will be greater
♦ If the strength of the electric field is lesser, the deflection (upward) will be lesser
• If SE denotes the strength of the electric field, we can write:
Upward deflection d ∝ SE
• Also we have:
♦ If the strength of the magnetic field is greater, the deflection (downward) will be greater
♦ If the strength of the magnetic field is lesser, the deflection (downward) will be lesser
• If SM denotes the strength of the magnetic field, we can write:
Downward deflection d ∝ SM
10. Combining the 3 results, we can write:
• Upward deflection $\mathbf\small{d_U\propto\frac{e\,S_E}{m_e}}$
• Downward deflection $\mathbf\small{d_D\propto\frac{e\,S_M}{m_e}}$
11. Consider the relation $\mathbf\small{d_U\propto\frac{e\,S_E}{m_e}}$
• This can be rearranged as $\mathbf\small{\frac{d_U}{S_E}\propto\frac{e}{m_e}}$
♦ Quantities on the left side are known quantities
♦ Quantities on the right side are unknown quantities
• Note that, e is in the numerator and me is in the denominator
12. Consider the relation $\mathbf\small{d_D\propto\frac{e\,S_M}{m_e}}$
• This can be rearranged as $\mathbf\small{\frac{d_U}{S_M}\propto\frac{e}{m_e}}$
♦ Quantities on the left side are known quantities
♦ Quantities on the right side are unknown quantities
• Note that in this case also, e is in the numerator and me is in the denominator
13. Thompson made accurate measurements of the deflections dU and dD
• He also noted down the strengths SE and SM in each case
• Using those measurements, he was able to calculate the $\mathbf\small{\frac{e}{m_e}}$ ratio
• The value he obtained was 1.758820 × 1011 C kg-1
• We will see the detailed mathematical calculations in higher classes
14. So we have: $\mathbf\small{\frac{e}{m_e}}$ = 1.758820 × 1011 C kg-1
• There are only two unknowns: e and me
• If any one of those two can be determined, the other can be easily calculated
• So our next aim is to find either e or me
♦ The upper plate is positively charged
♦ The lower plate is negatively charged
2. Small drops of oil are allowed to fall through a hole in the upper plate
• These drops initially do not have any charge
• But they are made to acquire charges
♦ For that, the air inside the apparatus is ionized using x rays
♦ The atoms of the air becomes ions
♦ These ions collide with the oil drops
♦ The oil drops thus become charged
3. Once the oil drops become charged, their fall can be controlled
♦ The oil drops can be made to fall slowly
♦ The oil drops can be made to fall quickly
♦ The oil drops can be even made to stay still
• Such control can be achieved by increasing or decreasing the voltage applied to the plates
• Such control is possible because, the negatively charged oil drops are attracted by the positively charged upper plate
4. The rate of fall of the drops is observed by a telescope
(‘Rate of fall’ means 'distance fallen in one second')
• The rate of fall and the applied voltage were carefully noted down
5. Based on those readings, the 'charge on the drops' were calculated. The results of the calculations are as follows:
• Some drops have a charge of 1.6 ×10-19 C
♦ This is 1 × 1.6 ×10-19
• Some drops have a charge of 3.2 ×10-19 C
♦ This is 2 × 1.6 ×10-19
• Some drops have a charge of 4.8 ×10-19 C
♦ This is 3 × 1.6 ×10-19
• Some drops have a charge of 6.4 ×10-19 C
♦ This is 4 × 1.6 ×10-19
6. So the charge is always an integral multiple of 1.6 ×10-19
■ We can conclude that the charge of one electron is 1.6 ×10-19
• Drops which acquire only one electron will have a charge of 1.6 ×10-19
• Drops which acquire two electrons will have a charge of (2 × 1.6 ×10-19) C
• Drops which acquire three electrons will have a charge of (3 × 1.6 ×10-19) C
So on . . .
■ This experiment was performed in 1909 by Robert Millikan (1868–1953) of the University of Chicago. The experiment came to be known as Millikan's Oil drop method
7. Electrons have a negative charge
• So we can write:
Using Millikan’s oil drop method, the charge of an electron is found to be e = -1.6 ×10-19 C
• The present accepted value is e = -1.602176 ×10-19 C
(i) $\mathbf\small{\frac{e}{m_e}}$ = 1.758820 × 1011 C kg-1
(ii) e = -1.602176 ×10-19 C
■ So we get:
$\mathbf\small{m_e=\frac{1.602176\times 10^{-19}\,\rm{C}}{1.758820\times 10^{11}\,\rm{C\,kg^{-1}}}= 9.1094\times 10^{-31}\,\rm{kg}}$
• Thus scientists successfully determined both e and me of the electron
1. The cathode rays start from cathode and move towards the anode
2. These are not visible rays
• Their presence can be detected using fluorescent or phosphorescent materials
3. This is just like creating pictures on a television screen
• Television picture tubes are cathode ray tubes coated with fluorescent or phosphorescent materials
• Different portions of the coating glow with different intensities when cathode rays fall on those portions with different intensities. This creates the required pictures on the screen
4. The cathode rays travel in a straight line if there are no electric or magnetic fields
5. In the presence of electric or magnetic fields, the cathode rays behave as if they are made up of negatively charged particles. We will see those details soon
• This helped scientists to conclude that, cathode rays consist of negatively charged particles
• They called those particles: electrons
6. The characteristics of the cathode rays do not depend upon the materials of the electrodes
• That is., even if electrodes made of different materials are used in the tube, the results of the experiment do not change
7. The characteristics of the cathode rays do not depend upon the gas present in the tube
• That is., even if any other gas is used in the tube, the results of the experiment do not change
8. The findings in (6) and (7) helped scientists to conclude that, nature of electrons do not depend on the type of materials
• In other words, electrons present in all materials have the same characteristics
Charge to mass ratio of electrons
1. The fig.2.2 (a) below shows a schematic diagram |
| Fig.2.2 (a) |
• It starts from the cathode
• And passes through the hole in the anode
2. The ray flows along the x direction
(The 3 mutually perpendicular axes are shown at right side bottom of the diagram)
3. Two electrodes (shown in cyan color) are placed in the path of the ray
• When a voltage is applied across these electrodes, an electric field is created between them
• The electrodes are arranged in such a way that, the electric field is perpendicular to the electron flow
• From the fig., it is clear that, the electric field is in the z direction
4. In the fig., no voltage is applied and so there is no electric field
• Since there is no electric field, the ray passes without any deviation and hits the point B on the fluorescent screen
5. Let us see what happens when voltage is applied:
• In fig.2.2 (b) below, the upper electrode is connected to the positive terminal of the battery and the lower electrode to the negative terminal
 |
| Fig.2.2 (b) |
6. In fig.2.2 (c) below, the voltage across the cyan electrodes is cut off. Instead, a magnetic field is applied
• The north pole of the magnet is shown in red color
• The south pole of the magnet is shown in blue color
• The poles are arranged in such a way that, the magnetic field is perpendicular to the electron flow
• From the fig., it is clear that, the magnetic field is in the y direction
 |
| Fig.2.2 (c) |
7. The points A, B and C lie in the same vertical line
• Let us see the reason for 'C' being in the same vertical line as A and B:
(i) We have learnt about Fleming's right hand rule in our earlier classes (Details here)
• Consider the fig.2.2 (d) below. It shows the right arm in an inverted position
 |
| Fig.2.2 (d) |
• The middle finger points in the direction of the current (brown arrow in the x direction)
(iii) So the thump points in the direction of the 'deflection of the current carrying conductor' (cyan arrow in the downward z direction)
• Thus we see how point C comes vertically below A and B
8. Now we get back to the main discussion
• We saw these:
♦ When the electric field is applied, the ray deviates upwards
♦ When the magnetic field is applied, the ray deviates downwards
■ So let us see what happens when both the fields are applied simultaneously
• The electric field tries to pull the ray upwards
• But this pulling effect can be nullified by applying a magnetic field of suitable strength
• When this is done, the ray passes without any deviation and hits the point B
• This is shown in fig. e below:
 |
| Fig.2.2 (e) |
• Three points were noted by him regarding the 'amount of deviation' from the normal path:
(i) The amount of deviation depends upon the magnitude of the charge of the particle
(ii) The amount of deviation depends upon the mass of the particle
(iii) The amount of deviation depends upon the strength of the fields
We will now see each of the 3 items in more detail
(i) Magnitude of the charge:
• We saw in fig. 2.2(a) that, when the electric field is applied, the ray deflects towards the positive electrode
• So we can conclude that, the particles are negatively charged
♦ If the magnitude of this negative charge is greater, the deflection will be greater
♦ If the magnitude of this negative charge is lesser, the deflection will be lesser
• So deflection (d) is directly proportional to the charge e
• We can write: d ∝ e
(The symbol '∝' represents 'is proportional to')
(ii) Mass of the particle
♦ If the mass is greater, deflection will be lesser
♦ If the mass is lesser, deflection will be greater
• So deflection (d) is inversely proportional to the mass m
♦ Where me is the mass of one electron
• We can write: $\mathbf\small{d\propto\frac{1}{m_e}}$
(iii) Strength of fields
♦ If the strength of the electric field is greater, the deflection (upward) will be greater
♦ If the strength of the electric field is lesser, the deflection (upward) will be lesser
• If SE denotes the strength of the electric field, we can write:
Upward deflection d ∝ SE
• Also we have:
♦ If the strength of the magnetic field is greater, the deflection (downward) will be greater
♦ If the strength of the magnetic field is lesser, the deflection (downward) will be lesser
• If SM denotes the strength of the magnetic field, we can write:
Downward deflection d ∝ SM
10. Combining the 3 results, we can write:
• Upward deflection $\mathbf\small{d_U\propto\frac{e\,S_E}{m_e}}$
• Downward deflection $\mathbf\small{d_D\propto\frac{e\,S_M}{m_e}}$
11. Consider the relation $\mathbf\small{d_U\propto\frac{e\,S_E}{m_e}}$
• This can be rearranged as $\mathbf\small{\frac{d_U}{S_E}\propto\frac{e}{m_e}}$
♦ Quantities on the left side are known quantities
♦ Quantities on the right side are unknown quantities
• Note that, e is in the numerator and me is in the denominator
12. Consider the relation $\mathbf\small{d_D\propto\frac{e\,S_M}{m_e}}$
• This can be rearranged as $\mathbf\small{\frac{d_U}{S_M}\propto\frac{e}{m_e}}$
♦ Quantities on the left side are known quantities
♦ Quantities on the right side are unknown quantities
• Note that in this case also, e is in the numerator and me is in the denominator
13. Thompson made accurate measurements of the deflections dU and dD
• He also noted down the strengths SE and SM in each case
• Using those measurements, he was able to calculate the $\mathbf\small{\frac{e}{m_e}}$ ratio
• The value he obtained was 1.758820 × 1011 C kg-1
• We will see the detailed mathematical calculations in higher classes
14. So we have: $\mathbf\small{\frac{e}{m_e}}$ = 1.758820 × 1011 C kg-1
• There are only two unknowns: e and me
• If any one of those two can be determined, the other can be easily calculated
• So our next aim is to find either e or me
Charge on the electron
1. In the schematic diagram in fig.2.3 below, two electrically charged plates are shown♦ The upper plate is positively charged
♦ The lower plate is negatively charged
 |
| Fig.2.3 |
• These drops initially do not have any charge
• But they are made to acquire charges
♦ For that, the air inside the apparatus is ionized using x rays
♦ The atoms of the air becomes ions
♦ These ions collide with the oil drops
♦ The oil drops thus become charged
3. Once the oil drops become charged, their fall can be controlled
♦ The oil drops can be made to fall slowly
♦ The oil drops can be made to fall quickly
♦ The oil drops can be even made to stay still
• Such control can be achieved by increasing or decreasing the voltage applied to the plates
• Such control is possible because, the negatively charged oil drops are attracted by the positively charged upper plate
4. The rate of fall of the drops is observed by a telescope
(‘Rate of fall’ means 'distance fallen in one second')
• The rate of fall and the applied voltage were carefully noted down
5. Based on those readings, the 'charge on the drops' were calculated. The results of the calculations are as follows:
• Some drops have a charge of 1.6 ×10-19 C
♦ This is 1 × 1.6 ×10-19
• Some drops have a charge of 3.2 ×10-19 C
♦ This is 2 × 1.6 ×10-19
• Some drops have a charge of 4.8 ×10-19 C
♦ This is 3 × 1.6 ×10-19
• Some drops have a charge of 6.4 ×10-19 C
♦ This is 4 × 1.6 ×10-19
6. So the charge is always an integral multiple of 1.6 ×10-19
■ We can conclude that the charge of one electron is 1.6 ×10-19
• Drops which acquire only one electron will have a charge of 1.6 ×10-19
• Drops which acquire two electrons will have a charge of (2 × 1.6 ×10-19) C
• Drops which acquire three electrons will have a charge of (3 × 1.6 ×10-19) C
So on . . .
■ This experiment was performed in 1909 by Robert Millikan (1868–1953) of the University of Chicago. The experiment came to be known as Millikan's Oil drop method
7. Electrons have a negative charge
• So we can write:
Using Millikan’s oil drop method, the charge of an electron is found to be e = -1.6 ×10-19 C
• The present accepted value is e = -1.602176 ×10-19 C
Now we can combine the two results:
(ii) e = -1.602176 ×10-19 C
■ So we get:
$\mathbf\small{m_e=\frac{1.602176\times 10^{-19}\,\rm{C}}{1.758820\times 10^{11}\,\rm{C\,kg^{-1}}}= 9.1094\times 10^{-31}\,\rm{kg}}$
• Thus scientists successfully determined both e and me of the electron
In the next section, we will see proton
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