We want to learn about the Bohr model of atom. For that, we had to 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 section 2.6, we saw how 'particle nature' came to be known to the scientific community
• In the section 2.7, we saw how Albert Einstein proved the 'particle nature'. We also saw that the 'concept of dual nature' was accepted by the scientific community
■ The next step was to learn about Quantization. In this regard,
• In the section 2.8, we saw the basics about atomic spectrum
• In the previous section 2.9, we saw the atomic spectrum of hydrogen
While learning about the hydrogen spectrum, several questions arise in our minds
Some sample questions are given below:
• The electron of hydrogen absorbed a ‘radiation of wavelength 656.47 nm’
♦ Why exactly 656.47 ?
♦ Why not 655 or 658 ?
• Why is there a large gap between the first right line and the second right line in all the series?
• Why is there very little gaps in the left most lines in all the series ?
All such questions were answered when Neils Bohr presented his model
In this section we will learn about his model in detail
• A postulate is something which is 'considered to be true'
♦ It can be considered to be true because, it gives explanations to the experimental results
♦ Also it provides basis for further discussions and developments
• Studies made by Neils Bohr enabled him to put forward 5 postulates regarding the structure of hydrogen atom
♦ Bohr arrived at the postulates through complex mathematical calculations
♦ At present we need not learn about those calculations
♦ What is important for us right now, is to get a good understanding about those postulates
Let us see the five postulates in detail. They are numbered from A to E:
Postulate A
1. The electron in the hydrogen atom moves around the nucleus in circular paths
• Note that, the paths are circular. Not elliptical
2. These paths are called orbits
• These paths are also called stationary states
• These paths are also called allowed energy states
3. So we can visualize some circles around the nucleus
• These circles are concentric. That is., all the circles will have the same center
• Naturally, the center will be the nucleus
This is shown in fig.2.27(a) below:
4. Consider any one of those circles
• It will be having a fixed radius
♦ That means, the radius of orbits will not change under any circumstances
5. Consider any one of those circles
• The electron moving along that circle will have a definite energy
♦ That energy will not change under any circumstances
Postulate B
1. The energy of an electron does not change if it remains in a orbit
2. But if that electron absorb energy, it will jump to a higher orbit
• This is shown in figs 2.27 (b) and (c)
• ‘Higher orbit’ means, an orbit having a higher energy level
3. Also, if that electron emit energy, it will jump to a lower orbit
• This is shown in figs 2.28 (a) and (b) below:
• ‘Lower orbit’ means, an orbit having a lower energy level
4. So it is clear that, an electron cannot absorb ‘any energy’
• The electron will first analyze all the incoming radiations
• It will absorb only that radiation which will enable it to land exactly on a higher orbit
5. If it absorb ‘any radiation’, it may end up in between two higher energy levels
• This never happens because, the electron never absorbs any unwanted radiations
• That is why, we see only definite values like 656.47 nm (the reddish line in Balmer series)
■ In general, we can write:
The electrons will absorb (and hence emit) only suitable radiations. That is why we see 'only lines' and 'no regions' in the atomic spectra
Postulate C
1. Each orbit has it’s own fixed energy level
2. If we can find the energy level of each of those orbits, we can find how much energy will be absorbed or emitted by the electrons
• This can be elaborated as follows:
(i) Let the electron be residing in an orbit
♦ That electron will be having an energy corresponding to that orbit
♦ We will call it the ‘initial energy’. We will denote that initial energy as Ei
(ii) Let that electron jump (by absorbing or emitting radiation) to a second orbit
♦ After the jump, it will be having an energy corresponding to the new orbit
♦ We will call it the ‘final energy’. We will denote that final energy as Ef
(iii) So the difference between the two energy levels will be Ef - Ei
♦ We will denote this difference as ΔE
♦ So we get: ΔE = Ef - Ei
(iv) We can write:
♦ ΔE is the difference between the two energy levels
♦ ΔE is also the energy absorbed/emitted by the electron while making the jump
(v) Let us see an example:
♦ Let initially, the electron be residing in the second orbit. Then Ei = E2
♦ Let that electron absorb energy and jump to the fourth orbit. Then Ef = E4
♦ So we get: The ‘quantity of energy’ absorbed = ΔE = Ef-Ei = E4-E2
(vi) Another example:
♦ Let initially, the electron be residing in the fifth orbit. Then Ei = E5
♦ Let that electron emit energy and jump to the third orbit. Then Ef = E3
♦ So we get: The ‘quantity of energy’ released = ΔE = Ef-Ei = E3-E5
3. We have seen that, energy is emitted or absorbed in packets (photons)
• Each photon has an energy of h𝝂 (Details here)
• So we can write: h𝝂 = ΔE = (Ef-Ei)
• From this we get: $\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{(E_f-E_i)}{h}}$
■ This expression is known as Bohr’s frequency rule
4. So, if we know the energy difference between two orbits, we can find the frequency of the radiation that is absorbed/emitted when the electron jumps between those two orbits
Postulate D
1. Consider a body moving in a linear path (straight line). We know that, such a body will have a linear momentum
• This momentum is given by: p = mv
• Where:
♦ p is the linear momentum
♦ m is the mass of the body
♦ v is the velocity of the body
2. If the body is moving in a curved path, it will have an angular momentum
• This angular momentum is given by: l = mvr
• Where:
♦ l is the angular momentum
♦ m is the mass of the body
♦ v is the linear velocity of the body
♦ r is the radius of the curved path
• We will see the derivation of this equation in physics classes
3. So, for an electron moving in a circular orbit, the angular momentum will be given by: le = mever
Where
♦ le is the angular momentum of the electron
♦ me is the mass of the electron
♦ ve is the linear velocity of the electron
♦ r is the radius of the circular orbit
4. Bohr found out that, this angular momentum will always be an integral multiple of $\mathbf\small{\frac{h}{2\pi}}$
• So we can write: $\mathbf\small{m_ev_er=\frac{nh}{2\pi}}$
• Where
♦ h is the Planck’s constant
♦ n = 1, 2, 3, . . .
5. The electron of hydrogen should satisfy the equation in (4)
• Since ‘n’ cannot be a fraction or decimal, ‘only certain orbits’ will be possible
6. Those 'orbits which are possible' are numbered as 1, 2, 3, . . .
• These integral numbers are called Principal quantum numbers
• So we have:
♦ The orbit with principal quantum number 1
♦ The orbit with principal quantum number 2
♦ The orbit with principal quantum number 3
♦ so on . . .
7. Bohr found out that, the radius of any orbit can be calculated using the expression: rn = n2a0
• Where:
♦ rn is the radius of the orbit with principal quantum number n
♦ a0 is a constant whose value is 52.9 pm
• Thus we get:
radius of the first orbit = r1 = 12 × 52.9 pm = 52.9 pm
• Note that, when n increases, r increases. That means, when n increases, the electron will be present more and more away from the nucleus
Postulate E
1. The most important information that we get from the Bohr’s model is the ‘energy associated with each orbit’
2. Bohr found out that:
The energy associated with an orbit is given by $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• Where:
♦ n is the principal quantum number of that orbit
♦ Rn is called the Rydberg constant whose value is 2.18×10-18 J
3. Thus we get:
• Energy associated with the first orbit
= E1 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{1^2}\right)}$ = -2.18×10-18 J
• Energy associated with the second orbit
= E2 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{2^2}\right)}$ = -0.545×10-18 J
• Energy associated with the third orbit
= E3 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{3^2}\right)}$ = -0.242×10-18 J
In the same way we get:
• E4 = -0.136×10-18 J
• E5 = -0.087×10-18 J
• E6 = -0.061×10-18 J
4. These energy levels are plotted on a graph in fig.2.29 below. Such a graph is called an energy level diagram
An example:
• For n = 2, we have: E2 = -0.545×10-18 J
♦ This value falls between -0.75 and -0.5
♦ So the energy E2 is represented by the second (counting upwards from bottom) yellow line in the fig.2.29
5. Significance of negative energy:
• We see that, there is a ‘-’ sign before each of the energy values. Why is that so ?
The answer can be written in 6 steps:
(i) Consider the energy at the lowest level (n=1). It is -2.18×10-18 J
• Energy at the next higher level (n=2) is -0.545×10-18 J
(ii) Bohr found out that, when the distance from the nucleus increases, energy of the electron increases
• That is., when n increases, the energy of the electron increases
(iii) We know that, when we compare -2.18×10-18 and -0.545×10-18, the larger value is -0.545×10-18
• This is a peculiarity of ‘negative numbers’ We learnt those details when we saw ‘number lines’
(iv) A number line is shown in fig.2.30 below:
• We can start from any point on the number line
♦ From that starting point, if we move towards the right, the values will be increasing
♦ From that starting point, if we move towards the left, the values will be decreasing
(v) So in fig.2.29, the energy in fact increases as n increases
• Thus we see that ‘-’ sign is essential for Bohr’s model to work
(vi) While doing the mathematical calculations, Bohr indeed obtained the ‘-’ sign
• We already saw that, the expression given by him is $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• So we can apply the ‘-’ sign with confidence
6. Now the next question arises:
If energy increases with n, what is the maximum energy possible ?
• The answer can be written in 4 steps:
(i) In the number line, consider the portion to the left of the zero
• If we start from a ‘point on the far left’, and move towards the right, the values will be increasing
• That is., values will be becoming less and less negative
• Finally, we will reach the zero at the middle
• That means, zero is greater than any negative value
(ii) In our present case, the expression is: $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• We put higher and higher values for n
(iii) Since n is in the denominator, En will become less and less negative
• It will become less and less negative and reach zero
(iv) For what value of n, does En become zero ?
• This is a question related to maths. The values allowable for n are: 1, 2, 3, . . .
• The maximum value is ‘infinity’ Infinity is represented as $\mathbf\small{\infty }$
• So we can write: Maximum value of En = $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty ^2}\right)}$
• But $\mathbf\small{\infty ^2}$ is $\mathbf\small{\infty}$
• So we get: $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty}\right)}$
• Consider $\mathbf\small{ \left(\frac{1}{\infty}\right)}$. We are dividing 1 by a very large number
• The result will be very very small. It will be very close to zero
• It can be considered to be zero for all practical purposes
• So we get: $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty}\right)=-R_H \times 0=0}$
Thus, the maximum energy possible is zero
7. We can write:
• If the electron is at a large distance away from the nucleus, the energy of that electron will be zero
• In such a situation, that electron will not be bound to the atom. It will be a ‘free electron’
• The hydrogen atom which has lost electron is said to be ionized
• We know that, a normal hydrogen atom has only one electron
• So an ionized hydrogen atom will not have any electron
8. Now we can really appreciate the importance of the ‘-’ sign in the expression $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• We have to satisfy two conditions:
(i) Energy must increase with the increase in ‘n’
(ii) Energy must become zero when n is $\mathbf\small{\infty }$
• If there is no ‘-’ sign, condition (ii) will be satisfied. But condition (i) will not be satisfied
• In section 2.5, we saw the 'wave nature of the electromagnetic radiations' and the 'electromagnetic spectrum'
• In section 2.6, we saw how 'particle nature' came to be known to the scientific community
• In the section 2.7, we saw how Albert Einstein proved the 'particle nature'. We also saw that the 'concept of dual nature' was accepted by the scientific community
■ The next step was to learn about Quantization. In this regard,
• In the section 2.8, we saw the basics about atomic spectrum
• In the previous section 2.9, we saw the atomic spectrum of hydrogen
While learning about the hydrogen spectrum, several questions arise in our minds
Some sample questions are given below:
• The electron of hydrogen absorbed a ‘radiation of wavelength 656.47 nm’
♦ Why exactly 656.47 ?
♦ Why not 655 or 658 ?
• Why is there a large gap between the first right line and the second right line in all the series?
• Why is there very little gaps in the left most lines in all the series ?
All such questions were answered when Neils Bohr presented his model
In this section we will learn about his model in detail
• A postulate is something which is 'considered to be true'
♦ It can be considered to be true because, it gives explanations to the experimental results
♦ Also it provides basis for further discussions and developments
• Studies made by Neils Bohr enabled him to put forward 5 postulates regarding the structure of hydrogen atom
♦ Bohr arrived at the postulates through complex mathematical calculations
♦ At present we need not learn about those calculations
♦ What is important for us right now, is to get a good understanding about those postulates
Let us see the five postulates in detail. They are numbered from A to E:
Postulate A
1. The electron in the hydrogen atom moves around the nucleus in circular paths
• Note that, the paths are circular. Not elliptical
2. These paths are called orbits
• These paths are also called stationary states
• These paths are also called allowed energy states
3. So we can visualize some circles around the nucleus
• These circles are concentric. That is., all the circles will have the same center
• Naturally, the center will be the nucleus
This is shown in fig.2.27(a) below:
 |
| Fig.2.27 |
• It will be having a fixed radius
♦ That means, the radius of orbits will not change under any circumstances
5. Consider any one of those circles
• The electron moving along that circle will have a definite energy
♦ That energy will not change under any circumstances
Postulate B
1. The energy of an electron does not change if it remains in a orbit
2. But if that electron absorb energy, it will jump to a higher orbit
• This is shown in figs 2.27 (b) and (c)
• ‘Higher orbit’ means, an orbit having a higher energy level
3. Also, if that electron emit energy, it will jump to a lower orbit
• This is shown in figs 2.28 (a) and (b) below:
 |
| Fig.2.28 |
4. So it is clear that, an electron cannot absorb ‘any energy’
• The electron will first analyze all the incoming radiations
• It will absorb only that radiation which will enable it to land exactly on a higher orbit
5. If it absorb ‘any radiation’, it may end up in between two higher energy levels
• This never happens because, the electron never absorbs any unwanted radiations
• That is why, we see only definite values like 656.47 nm (the reddish line in Balmer series)
■ In general, we can write:
The electrons will absorb (and hence emit) only suitable radiations. That is why we see 'only lines' and 'no regions' in the atomic spectra
Postulate C
1. Each orbit has it’s own fixed energy level
2. If we can find the energy level of each of those orbits, we can find how much energy will be absorbed or emitted by the electrons
• This can be elaborated as follows:
(i) Let the electron be residing in an orbit
♦ That electron will be having an energy corresponding to that orbit
♦ We will call it the ‘initial energy’. We will denote that initial energy as Ei
(ii) Let that electron jump (by absorbing or emitting radiation) to a second orbit
♦ After the jump, it will be having an energy corresponding to the new orbit
♦ We will call it the ‘final energy’. We will denote that final energy as Ef
(iii) So the difference between the two energy levels will be Ef - Ei
♦ We will denote this difference as ΔE
♦ So we get: ΔE = Ef - Ei
(iv) We can write:
♦ ΔE is the difference between the two energy levels
♦ ΔE is also the energy absorbed/emitted by the electron while making the jump
(v) Let us see an example:
♦ Let initially, the electron be residing in the second orbit. Then Ei = E2
♦ Let that electron absorb energy and jump to the fourth orbit. Then Ef = E4
♦ So we get: The ‘quantity of energy’ absorbed = ΔE = Ef-Ei = E4-E2
(vi) Another example:
♦ Let initially, the electron be residing in the fifth orbit. Then Ei = E5
♦ Let that electron emit energy and jump to the third orbit. Then Ef = E3
♦ So we get: The ‘quantity of energy’ released = ΔE = Ef-Ei = E3-E5
3. We have seen that, energy is emitted or absorbed in packets (photons)
• Each photon has an energy of h𝝂 (Details here)
• So we can write: h𝝂 = ΔE = (Ef-Ei)
• From this we get: $\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{(E_f-E_i)}{h}}$
■ This expression is known as Bohr’s frequency rule
4. So, if we know the energy difference between two orbits, we can find the frequency of the radiation that is absorbed/emitted when the electron jumps between those two orbits
Postulate D
1. Consider a body moving in a linear path (straight line). We know that, such a body will have a linear momentum
• This momentum is given by: p = mv
• Where:
♦ p is the linear momentum
♦ m is the mass of the body
♦ v is the velocity of the body
2. If the body is moving in a curved path, it will have an angular momentum
• This angular momentum is given by: l = mvr
• Where:
♦ l is the angular momentum
♦ m is the mass of the body
♦ v is the linear velocity of the body
♦ r is the radius of the curved path
• We will see the derivation of this equation in physics classes
3. So, for an electron moving in a circular orbit, the angular momentum will be given by: le = mever
Where
♦ le is the angular momentum of the electron
♦ me is the mass of the electron
♦ ve is the linear velocity of the electron
♦ r is the radius of the circular orbit
4. Bohr found out that, this angular momentum will always be an integral multiple of $\mathbf\small{\frac{h}{2\pi}}$
• So we can write: $\mathbf\small{m_ev_er=\frac{nh}{2\pi}}$
• Where
♦ h is the Planck’s constant
♦ n = 1, 2, 3, . . .
5. The electron of hydrogen should satisfy the equation in (4)
• Since ‘n’ cannot be a fraction or decimal, ‘only certain orbits’ will be possible
6. Those 'orbits which are possible' are numbered as 1, 2, 3, . . .
• These integral numbers are called Principal quantum numbers
• So we have:
♦ The orbit with principal quantum number 1
♦ The orbit with principal quantum number 2
♦ The orbit with principal quantum number 3
♦ so on . . .
7. Bohr found out that, the radius of any orbit can be calculated using the expression: rn = n2a0
• Where:
♦ rn is the radius of the orbit with principal quantum number n
♦ a0 is a constant whose value is 52.9 pm
• Thus we get:
radius of the first orbit = r1 = 12 × 52.9 pm = 52.9 pm
• Note that, when n increases, r increases. That means, when n increases, the electron will be present more and more away from the nucleus
Postulate E
1. The most important information that we get from the Bohr’s model is the ‘energy associated with each orbit’
2. Bohr found out that:
The energy associated with an orbit is given by $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• Where:
♦ n is the principal quantum number of that orbit
♦ Rn is called the Rydberg constant whose value is 2.18×10-18 J
3. Thus we get:
• Energy associated with the first orbit
= E1 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{1^2}\right)}$ = -2.18×10-18 J
• Energy associated with the second orbit
= E2 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{2^2}\right)}$ = -0.545×10-18 J
• Energy associated with the third orbit
= E3 = $\mathbf\small{E_1=-2.18\times10^{-18}\times \left(\frac{1}{3^2}\right)}$ = -0.242×10-18 J
In the same way we get:
• E4 = -0.136×10-18 J
• E5 = -0.087×10-18 J
• E6 = -0.061×10-18 J
4. These energy levels are plotted on a graph in fig.2.29 below. Such a graph is called an energy level diagram
 |
| Fig.2.29 |
• For n = 2, we have: E2 = -0.545×10-18 J
♦ This value falls between -0.75 and -0.5
♦ So the energy E2 is represented by the second (counting upwards from bottom) yellow line in the fig.2.29
5. Significance of negative energy:
• We see that, there is a ‘-’ sign before each of the energy values. Why is that so ?
The answer can be written in 6 steps:
(i) Consider the energy at the lowest level (n=1). It is -2.18×10-18 J
• Energy at the next higher level (n=2) is -0.545×10-18 J
(ii) Bohr found out that, when the distance from the nucleus increases, energy of the electron increases
• That is., when n increases, the energy of the electron increases
(iii) We know that, when we compare -2.18×10-18 and -0.545×10-18, the larger value is -0.545×10-18
• This is a peculiarity of ‘negative numbers’ We learnt those details when we saw ‘number lines’
(iv) A number line is shown in fig.2.30 below:
 |
| Fig.2.30 |
♦ From that starting point, if we move towards the right, the values will be increasing
♦ From that starting point, if we move towards the left, the values will be decreasing
(v) So in fig.2.29, the energy in fact increases as n increases
• Thus we see that ‘-’ sign is essential for Bohr’s model to work
(vi) While doing the mathematical calculations, Bohr indeed obtained the ‘-’ sign
• We already saw that, the expression given by him is $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• So we can apply the ‘-’ sign with confidence
6. Now the next question arises:
If energy increases with n, what is the maximum energy possible ?
• The answer can be written in 4 steps:
(i) In the number line, consider the portion to the left of the zero
• If we start from a ‘point on the far left’, and move towards the right, the values will be increasing
• That is., values will be becoming less and less negative
• Finally, we will reach the zero at the middle
• That means, zero is greater than any negative value
(ii) In our present case, the expression is: $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• We put higher and higher values for n
(iii) Since n is in the denominator, En will become less and less negative
• It will become less and less negative and reach zero
(iv) For what value of n, does En become zero ?
• This is a question related to maths. The values allowable for n are: 1, 2, 3, . . .
• The maximum value is ‘infinity’ Infinity is represented as $\mathbf\small{\infty }$
• So we can write: Maximum value of En = $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty ^2}\right)}$
• But $\mathbf\small{\infty ^2}$ is $\mathbf\small{\infty}$
• So we get: $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty}\right)}$
• Consider $\mathbf\small{ \left(\frac{1}{\infty}\right)}$. We are dividing 1 by a very large number
• The result will be very very small. It will be very close to zero
• It can be considered to be zero for all practical purposes
• So we get: $\mathbf\small{E_{\infty }=-R_H \left(\frac{1}{\infty}\right)=-R_H \times 0=0}$
Thus, the maximum energy possible is zero
7. We can write:
• If the electron is at a large distance away from the nucleus, the energy of that electron will be zero
• In such a situation, that electron will not be bound to the atom. It will be a ‘free electron’
• The hydrogen atom which has lost electron is said to be ionized
• We know that, a normal hydrogen atom has only one electron
• So an ionized hydrogen atom will not have any electron
8. Now we can really appreciate the importance of the ‘-’ sign in the expression $\mathbf\small{E_n=-R_H \left(\frac{1}{n^2}\right)}$
• We have to satisfy two conditions:
(i) Energy must increase with the increase in ‘n’
(ii) Energy must become zero when n is $\mathbf\small{\infty }$
• If there is no ‘-’ sign, condition (ii) will be satisfied. But condition (i) will not be satisfied
This completes a discussion on the basics of Bohr model of hydrogen atom. In the next section we will see how the model explains the various lines in the hydrogen spectrum
No comments:
Post a Comment