In the previous section, we saw details about atomic number, mass number, isobars and isotopes. We also saw some solved examples. In this section we will see a few more solved examples. Later in this section, we will see some drawbacks of the Rutherford's model
Solved example 2.7
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across of scale of length 20 cm long
Solution:
1. One nm = 10-9 m
⇒ 0.15 nm = 0.15 × 10-9 m
2. 1 cm = 0.01 m
⇒ 20 cm = 20 × 0.01 = 0.2 m
3. Thus number of atoms = $\mathbf\small{\frac{0.2}{0.15\times10^{-9}}}$ = 1.33 × 109 numbers
Solved example 2.8
2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm
Solution:
• Let the diameter of the carbon atom be 'd' m
• Then we get: (2 ×108)d = 0.024 m
⇒ d = $\mathbf\small{\frac{0.024}{2\times10^{8}}}$ = 1.2 × 10-10 m
⇒ radius r = $\mathbf\small{\frac{1.2\times10^{-10}}{2}}$ = 6.0 × 10-11 m
Solved example 2.9
The diameter of zinc atom is 2.6 $\mathbf\small{\rm{\mathring{A}}}$. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise
Solution:
Part (a):
1. One $\mathbf\small{\rm{\mathring{A}}}$= 10-10 m
• So diameter = 2.6 × 10-10 m
⇒ Radius = 1.3 × 10-10 m
2. 1 pm = 10-12 m
⇒ 1 m = 1012 pm
3. So 1.3 × 10-10 m = (1.3 × 10-10) × 1012 = 130 pm
Part (b):
1. 1 cm = 0.01 m
⇒ 1.6 cm = 0.016 m
2. So number of zinc atoms = $\mathbf\small{\frac{0.016}{2.6\times10^{-10}}}$ = 6.15 × 107 numbers
Solved example 2.10
A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it
Solution:
• Let the number of electrons be x
• Charge of one electron = 1.602176 ×10-19 C
• So total charge of x electrons = (1.602176 ×10-19)x = 2.5 × 10-16
⇒ x = $\mathbf\small{\frac{2.5\times10^{-16}}{1.602176\times10^{-19}}}$ = 1560 numbers
Solved example 2.11
In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is -1.282 × 10-18 C, calculate the number of electrons present on it
Solution:
• Let the number of electrons be x
• Charge of one electron = -1.602176 ×10-19 C
• So total charge of x electrons = (-1.602176 ×10-19)x = –1.282 × 10-18
⇒ x = $\mathbf\small{\frac{–1.282\times10^{-18}}{1.602176\times10^{-19}}}$ = 8 numbers
Solved example 2.12
In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Solution:
1. In the experiment, some α-particles are bounced back (nearly 180o deflection)
• This 'bouncing back' is the evidence which shows the presence of 'some thing massive' inside the atom
(We saw the details in a previous section)
2. Gold atoms are heavy
• But even when a gold foil is used, we get only a very few (1 in 20000) 'large deflections'
3. If we use smaller atoms, most of the α-particles will suffer only a small deviation
• Because, the nucleus will have only a small mass and only a small positive charge to cause repulsion
• The results will not suggest the presence of 'some thing massive'
Solved example 2.13
Symbols $\mathbf\small{\rm{^{79}_{35}Br}}$ and $\mathbf\small{\rm{^{79}Br}}$ can be written, whereas symbols $\mathbf\small{\rm{^{35}_{79}Br}}$ and $\mathbf\small{\rm{^{35}Br}}$ are not acceptable. Answer briefly
Solution:
1. Br has the atomic number 35 and mass number 79
• As per IUPAC rules, mass number should be the superscript and atomic number should be the subscript. So $\mathbf\small{\rm{^{35}_{79}Br}}$ is not acceptable
2. $\mathbf\small{\rm{^{35}Br}}$ gives the impression that, 35 is the mass number of Br. So this is also not acceptable
Solved example 2.14
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol
Solution:
1. Given that A = 81
• Let number of protons be Z
• Let number of neutrons be n
2. Given that n = (1.317)Z
∵ n = Z + 31.7% of Z
3. We have: n+ Z = A
Substituting the values, we get:
(1.317)Z + Z = 81
⇒ Z = 34.959
⇒ Z ≈ 35
4. From the periodic table, we have Z = 35 for Br
• Using the notation $\mathbf\small{\rm{^A_ZX}}$, we get the atomic symbol as: $\mathbf\small{\rm{^{81}_{35}Br}}$
Solved example 2.15
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion
Solution:
1. Given that A = 37
• Let number of electrons be e
• Let number of protons be Z
• Let number of neutrons be n
2. Given that the ion has one unit of negative charge
• so e = (Z+1)
3. Given that n = (1.111)e
∵ n = e + 11.1% of e
4. Substituting (2) in (3), we get:
n = (1.111)(Z+1)
5. But Z = (A-n) = (37-n)
• Substituting this in (4), we get:
⇒ n = (1.111)[(37-n)+1]
⇒ n = (1.111)[38-n]
⇒ n = 19.999
⇒ n ≈ 20
6. So we get Z = (37-20) = 17
• From the periodic table, we have Z = 17 for Cl
7. So the symbol is: $\mathbf\small{\rm{^{37}_{17}Cl^{1-}}}$
Solved example 2.16
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion
Solution:
1. Given that A = 56
• Let number of electrons be e
• Let number of protons be Z
• Let number of neutrons be n
2. Given that the ion has three units of positive charge
• so Z = (e+3)
⇒ e = Z - 3
3. Given that n = (1.304)e
∵ n = e + 30.4% of e
4. Substituting (2) in (3), we get:
n = (1.304)(Z-3)
5. But Z = (A-n) = (56-n)
• Substituting this in (4), we get:
⇒ n = (1.304)[(56-n)-3]
⇒ n = (1.304)[53-n]
⇒ n = 29.99
⇒ n ≈ 30
6. So we get Z = (56-30) = 26
From the periodic table, we have Z = 26 for Fe
7. So the symbol is: $\mathbf\small{\rm{^{56}_{26}Fe^{3+}}}$
Solved example 2.7
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across of scale of length 20 cm long
Solution:
1. One nm = 10-9 m
⇒ 0.15 nm = 0.15 × 10-9 m
2. 1 cm = 0.01 m
⇒ 20 cm = 20 × 0.01 = 0.2 m
3. Thus number of atoms = $\mathbf\small{\frac{0.2}{0.15\times10^{-9}}}$ = 1.33 × 109 numbers
Solved example 2.8
2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm
Solution:
• Let the diameter of the carbon atom be 'd' m
• Then we get: (2 ×108)d = 0.024 m
⇒ d = $\mathbf\small{\frac{0.024}{2\times10^{8}}}$ = 1.2 × 10-10 m
⇒ radius r = $\mathbf\small{\frac{1.2\times10^{-10}}{2}}$ = 6.0 × 10-11 m
Solved example 2.9
The diameter of zinc atom is 2.6 $\mathbf\small{\rm{\mathring{A}}}$. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise
Solution:
Part (a):
1. One $\mathbf\small{\rm{\mathring{A}}}$= 10-10 m
• So diameter = 2.6 × 10-10 m
⇒ Radius = 1.3 × 10-10 m
2. 1 pm = 10-12 m
⇒ 1 m = 1012 pm
3. So 1.3 × 10-10 m = (1.3 × 10-10) × 1012 = 130 pm
Part (b):
1. 1 cm = 0.01 m
⇒ 1.6 cm = 0.016 m
2. So number of zinc atoms = $\mathbf\small{\frac{0.016}{2.6\times10^{-10}}}$ = 6.15 × 107 numbers
Solved example 2.10
A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it
Solution:
• Let the number of electrons be x
• Charge of one electron = 1.602176 ×10-19 C
• So total charge of x electrons = (1.602176 ×10-19)x = 2.5 × 10-16
⇒ x = $\mathbf\small{\frac{2.5\times10^{-16}}{1.602176\times10^{-19}}}$ = 1560 numbers
Solved example 2.11
In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is -1.282 × 10-18 C, calculate the number of electrons present on it
Solution:
• Let the number of electrons be x
• Charge of one electron = -1.602176 ×10-19 C
• So total charge of x electrons = (-1.602176 ×10-19)x = –1.282 × 10-18
⇒ x = $\mathbf\small{\frac{–1.282\times10^{-18}}{1.602176\times10^{-19}}}$ = 8 numbers
Solved example 2.12
In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Solution:
1. In the experiment, some α-particles are bounced back (nearly 180o deflection)
• This 'bouncing back' is the evidence which shows the presence of 'some thing massive' inside the atom
(We saw the details in a previous section)
2. Gold atoms are heavy
• But even when a gold foil is used, we get only a very few (1 in 20000) 'large deflections'
3. If we use smaller atoms, most of the α-particles will suffer only a small deviation
• Because, the nucleus will have only a small mass and only a small positive charge to cause repulsion
• The results will not suggest the presence of 'some thing massive'
Solved example 2.13
Symbols $\mathbf\small{\rm{^{79}_{35}Br}}$ and $\mathbf\small{\rm{^{79}Br}}$ can be written, whereas symbols $\mathbf\small{\rm{^{35}_{79}Br}}$ and $\mathbf\small{\rm{^{35}Br}}$ are not acceptable. Answer briefly
Solution:
1. Br has the atomic number 35 and mass number 79
• As per IUPAC rules, mass number should be the superscript and atomic number should be the subscript. So $\mathbf\small{\rm{^{35}_{79}Br}}$ is not acceptable
2. $\mathbf\small{\rm{^{35}Br}}$ gives the impression that, 35 is the mass number of Br. So this is also not acceptable
Solved example 2.14
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol
Solution:
1. Given that A = 81
• Let number of protons be Z
• Let number of neutrons be n
2. Given that n = (1.317)Z
∵ n = Z + 31.7% of Z
3. We have: n+ Z = A
Substituting the values, we get:
(1.317)Z + Z = 81
⇒ Z = 34.959
⇒ Z ≈ 35
4. From the periodic table, we have Z = 35 for Br
• Using the notation $\mathbf\small{\rm{^A_ZX}}$, we get the atomic symbol as: $\mathbf\small{\rm{^{81}_{35}Br}}$
Solved example 2.15
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion
Solution:
1. Given that A = 37
• Let number of electrons be e
• Let number of protons be Z
• Let number of neutrons be n
2. Given that the ion has one unit of negative charge
• so e = (Z+1)
3. Given that n = (1.111)e
∵ n = e + 11.1% of e
4. Substituting (2) in (3), we get:
n = (1.111)(Z+1)
5. But Z = (A-n) = (37-n)
• Substituting this in (4), we get:
⇒ n = (1.111)[(37-n)+1]
⇒ n = (1.111)[38-n]
⇒ n = 19.999
⇒ n ≈ 20
6. So we get Z = (37-20) = 17
• From the periodic table, we have Z = 17 for Cl
7. So the symbol is: $\mathbf\small{\rm{^{37}_{17}Cl^{1-}}}$
Solved example 2.16
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion
Solution:
1. Given that A = 56
• Let number of electrons be e
• Let number of protons be Z
• Let number of neutrons be n
2. Given that the ion has three units of positive charge
• so Z = (e+3)
⇒ e = Z - 3
3. Given that n = (1.304)e
∵ n = e + 30.4% of e
4. Substituting (2) in (3), we get:
n = (1.304)(Z-3)
5. But Z = (A-n) = (56-n)
• Substituting this in (4), we get:
⇒ n = (1.304)[(56-n)-3]
⇒ n = (1.304)[53-n]
⇒ n = 29.99
⇒ n ≈ 30
6. So we get Z = (56-30) = 26
From the periodic table, we have Z = 26 for Fe
7. So the symbol is: $\mathbf\small{\rm{^{56}_{26}Fe^{3+}}}$
Now we will discuss about the drawbacks of Rutherford's model
1. We saw that, the Rutherford’s model resembles the solar system
• The nucleus plays the role of the sun
• The electrons play the role of the planets
2. In the solar system, the gravitational force of attraction between the bodies is given by $\mathbf\small{G\frac{m_1m_2}{r^2}}$
• Where:
♦ G is the gravitational constant
♦ m1 and m2 are the masses of the two bodies
♦ r is the distance between the two bodies
3. The force of attraction or repulsion between two charged particles is given by $\mathbf\small{k\frac{q_1q_2}{r^2}}$
• Where
♦ k is the Coulomb's constant
♦ q1 and q2 are the charges of the two particles
♦ r is the distance between the two particles
4. The equations in (2) and (3) are similar
(i) When the equation in (2) is applied to the planets in the solar system, we get the well defined paths
♦ These paths are used by those planets to move around the sun
(ii) That means, we can theoretically calculate the orbits of the planets
(iii) Also, the actual orbits used by those planets can be determined by making astronomical observations
(iv) The actual orbits thus determined are indeed same as those calculated theoretically
♦ This proves that the theoretical calculations are correct
5. Now consider the electrons moving around the nucleus
• The equations in (2) and (3) are similar
• So we would expect the electrons also to move in well defined orbits around the nucleus
6. But there is a problem. It can be written in 10 steps:
(i) ‘Motion in an orbit’ is different from ‘motion in a straight line’
• If a body moves in a straight line with uniform speed, we say that, the body has constant velocity
(ii) But in the case of 'motion in an orbit', the direction of motion changes continuously
• So, even if the speed is uniform, velocity is not constant
(iii) If the velocity is not constant, there must be acceleration
• That is, a body or a particle moving in an orbit will be experiencing acceleration, even if the speed is uniform
(iv) The planets moving around the sun will be experiencing acceleration
• Electrons moving around the nucleus will also be experiencing acceleration
(v) If a charged particle is subjected to acceleration, it will emit electromagnetic radiations
• The planets do not emit radiations because, they are not charged
(vi) Electromagnetic radiation is a form of energy
• So, if a charged particle is emitting this radiation, it means that, the particle is giving off energy
(vii) This energy is made available from it’s charge
• When it gives off energy, the charge continuously decreases
• If the charge decreases, the distance r decreases
(viii) Continuous decrease in charge will cause continuous decrease in distance
• So the orbit will not be a circle. Because, the radius continuously decreases
(ix) Such an orbit will be a spiral. This is shown in fig.2.7 below:
• The spiral ends in the nucleus
(x) So theoretically, the electron should follow the spiral path and end up inside the nucleus
• But this does not happen in real life
7. Now another question arises in our minds:
■The electron emits radiation because it moves around in an orbit. Then why not consider the electron to be stationary at a point away from the nucleus?
• The explanation can be written in 7 steps:
(i) The electron is stationary
(ii) The nucleus is also stationary
(iii) The nucleus is more massive than the electron
• The greater mass is due to the concentration of all the protons and neutrons inside the nucleus
(iv) Also the nucleus has greater positive charge
• This positive charge can be matched only if all the electrons stick together
(v) But the electrons do not stick together. They are distributed around the nucleus
• So the attraction is between the ‘heavy nucleus’ and the ‘light electron’
(vi) The electron will be easily pulled into the nucleus
• All the electrons will be pulled into the nucleus in this way
(vii) This type of pulling can be avoided only if the electron moves around the nucleus
8. So Rutherford’s model which predicts well defined orbits cannot be accepted
9. Another serious drawback:
• Rutherford’s model does not say any thing about the distribution of electrons
• That is., we want answers to the following questions:
(i) What is the distance between each electron and the nucleus ?
(ii) Do all orbits have the same radius ?
(iii) Do all electrons have the same energy ?
Rutherford’s model do not answer these questions
1. We saw that, the Rutherford’s model resembles the solar system
• The nucleus plays the role of the sun
• The electrons play the role of the planets
2. In the solar system, the gravitational force of attraction between the bodies is given by $\mathbf\small{G\frac{m_1m_2}{r^2}}$
• Where:
♦ G is the gravitational constant
♦ m1 and m2 are the masses of the two bodies
♦ r is the distance between the two bodies
3. The force of attraction or repulsion between two charged particles is given by $\mathbf\small{k\frac{q_1q_2}{r^2}}$
• Where
♦ k is the Coulomb's constant
♦ q1 and q2 are the charges of the two particles
♦ r is the distance between the two particles
4. The equations in (2) and (3) are similar
(i) When the equation in (2) is applied to the planets in the solar system, we get the well defined paths
♦ These paths are used by those planets to move around the sun
(ii) That means, we can theoretically calculate the orbits of the planets
(iii) Also, the actual orbits used by those planets can be determined by making astronomical observations
(iv) The actual orbits thus determined are indeed same as those calculated theoretically
♦ This proves that the theoretical calculations are correct
5. Now consider the electrons moving around the nucleus
• The equations in (2) and (3) are similar
• So we would expect the electrons also to move in well defined orbits around the nucleus
6. But there is a problem. It can be written in 10 steps:
(i) ‘Motion in an orbit’ is different from ‘motion in a straight line’
• If a body moves in a straight line with uniform speed, we say that, the body has constant velocity
(ii) But in the case of 'motion in an orbit', the direction of motion changes continuously
• So, even if the speed is uniform, velocity is not constant
(iii) If the velocity is not constant, there must be acceleration
• That is, a body or a particle moving in an orbit will be experiencing acceleration, even if the speed is uniform
(iv) The planets moving around the sun will be experiencing acceleration
• Electrons moving around the nucleus will also be experiencing acceleration
(v) If a charged particle is subjected to acceleration, it will emit electromagnetic radiations
• The planets do not emit radiations because, they are not charged
(vi) Electromagnetic radiation is a form of energy
• So, if a charged particle is emitting this radiation, it means that, the particle is giving off energy
(vii) This energy is made available from it’s charge
• When it gives off energy, the charge continuously decreases
• If the charge decreases, the distance r decreases
(viii) Continuous decrease in charge will cause continuous decrease in distance
• So the orbit will not be a circle. Because, the radius continuously decreases
(ix) Such an orbit will be a spiral. This is shown in fig.2.7 below:
 |
| Fig.2.7 |
(x) So theoretically, the electron should follow the spiral path and end up inside the nucleus
• But this does not happen in real life
7. Now another question arises in our minds:
■The electron emits radiation because it moves around in an orbit. Then why not consider the electron to be stationary at a point away from the nucleus?
• The explanation can be written in 7 steps:
(i) The electron is stationary
(ii) The nucleus is also stationary
(iii) The nucleus is more massive than the electron
• The greater mass is due to the concentration of all the protons and neutrons inside the nucleus
(iv) Also the nucleus has greater positive charge
• This positive charge can be matched only if all the electrons stick together
(v) But the electrons do not stick together. They are distributed around the nucleus
• So the attraction is between the ‘heavy nucleus’ and the ‘light electron’
(vi) The electron will be easily pulled into the nucleus
• All the electrons will be pulled into the nucleus in this way
(vii) This type of pulling can be avoided only if the electron moves around the nucleus
8. So Rutherford’s model which predicts well defined orbits cannot be accepted
9. Another serious drawback:
• Rutherford’s model does not say any thing about the distribution of electrons
• That is., we want answers to the following questions:
(i) What is the distance between each electron and the nucleus ?
(ii) Do all orbits have the same radius ?
(iii) Do all electrons have the same energy ?
Rutherford’s model do not answer these questions
In the next section we will see Bohr model of atom
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