In the previous section 2.10, we saw the basic details about Bohr model of hydrogen atom. In this section, we will see how the Bohr model can be used to explain the hydrogen spectrum
1. In the postulate C, we saw this: $\mathbf\small{\Delta E = E_f-E_i}$
2. But postulate E gives the exact values of Ei and Ef:
• $\mathbf\small{E_f=-R_H \left(\frac{1}{n_f^2}\right)}$
$\mathbf\small{\Delta E=\left[-R_H \left(\frac{1}{n_f^2}\right)--R_H \left(\frac{1}{n_i^2}\right)\right]}$
$\mathbf\small{\Rightarrow \Delta E=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
4. This expression in (3) gives the energy which is absorbed/emitted when the electron makes the jump. Let us apply it for absorption and emission separately:
■ If it is an absorption, the electron will be jumping from a lower state to a higher state
♦ In such a jump, nf > ni
♦ So the term inside the parenthesis will be positive
♦ Thus ΔE will be positive
■ If it is an emission, the electron will be jumping from a higher state to a lower state
♦ In such a jump, ni > nf
♦ So the term inside the parenthesis will be negative
♦ Thus ΔE will be negative
5. Thus we can write:
• A positive ΔE indicates absorption
• A negative ΔE indicates emission
6. Frequency of radiation:
• In the postulate C, we saw the Bohr’s frequency rule: $\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{(E_f-E_i)}{h}}$
• This rule gives the frequency of the radiation which is absorbed/emitted when the electron makes the jump
7. But postulate E gives the exact values of Ei and Ef:
• $\mathbf\small{E_f=-R_H \left(\frac{1}{n_f^2}\right)}$
$\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{1}{h}\left[-R_H \left(\frac{1}{n_f^2}\right)--R_H \left(\frac{1}{n_i^2}\right)\right]}$
$\mathbf\small{\Rightarrow \nu=\frac{R_H}{h}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
9. Substituting the values of RH and h in (8), we get:
$\mathbf\small{\nu=\frac{2.18\times 10^{-18}(J)}{6.626\times 10^{-34}(J s)}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
$\mathbf\small{\Rightarrow \nu=3.29\times 10^{15}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){s^{-1}}}$
• Thus we can calculate the frequency of the radiation absorbed/emitted when the electron makes the jump
10. Earlier, we saw the expression given by Rydberg: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)}$ (Details here)
• This expression is in terms of the 'wave number' $\mathbf\small{\bar{\nu}}$
• We have: $\mathbf\small{\nu=\frac{c}{\lambda}}$
$\mathbf\small{\Rightarrow \bar{\nu}=\frac{1}{\lambda}=\frac{\nu}{c}}$
• Thus our present equation in (9) becomes:
$\mathbf\small{\bar{\nu}=\frac{3.29\times 10^{15}}{3\times 10^{8}}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
$\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
■ This is the same equation derived by Rydberg
• However we must remember the two aspects:
(i) Bohr derived the equation by analyzing the internal structure of the hydrogen atom
(ii) Rydberg derived the equation with out any knowledge about the internal structure of the hydrogen atom. He used pure mathematics
12. Now we will see how Bohr’s model explains the lines seen in hydrogen spectrum
• We will use the lines in the Balmer series as an example
• We saw the Balmer series earlier in fig.2.23. (Details here) It is shown again below:
13. We saw that, the right most line represents a radiation of wave number 15232.9167 cm-1
(i) This wave number was derived using the expression given by Balmer: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{2^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 3
(ii) Now consider the expression in (11). It is the equation derived by Bohr: $\mathbf\small{\bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
The electron jumps from the third energy level (n=3) to the second energy level (n=2)
14. We saw that, the 'second line from the right' represents a radiation of wave number 20564.4375 cm-1
(i) This wave number was derived using the expression given by Balmer: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{2^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 4
(ii) Now consider the expression derived by Bohr: $\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
The electron jumps from the fourth energy level (n=4) to the second energy level (n=2)
15. In this way, all lines in the Balmer series can be explained
• We will see one more example. This time, from another series, say Paschen series
• We saw that, the right most line in the Paschen series represents a radiation of wave number 1.87564 × 10-4 cm-1
(i) This wave number was derived using the expression given by Paschen: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{3^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 4
(ii) Now consider the expression derived by Bohr: $\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
The electron jumps from the fourth energy level (n=4) to the third energy level (n=3)
• In this way, all lines in the Paschen series can be explained
16. Thus we can write:
It was Neils Bohr who solved the mystery of 'how the lines are formed in the hydrogen spectrum'
17. We saw three examples:
(i) In (13), we saw this:
• Right most line in the Balmer series:
♦ The electron jumps from the third energy level (n=3) to the second energy level (n=2)
(ii) In (14), we saw this:
• Second line from the right in the Balmer series:
♦ The electron jumps from the fourth energy level (n=4) to the second energy level (n=2)
(iii) In (15), we saw this:
• Right most line in the Paschen series:
♦ The electron jumps from the third energy level (n=4) to the second energy level (n=3)
• We will mark these three 'jumps' in the energy level diagram that we saw in fig.2.29 in the previous section. The modified fig. is shown in fig.2.31 below:
18. In this way, we can mark the jumps corresponding to all the other lines in the various series
• The jumps corresponding to Lyman, Blamer and Paschen series are shown in fig.2.32 below:
19. Thus we can write:
• Each line in the hydrogen spectrum (absorption spectrum as well as emission spectrum) can be related to 'a jump between two energy levels'
• Some of the jumps giving emission spectrum are:
(n=6) to (n=5), (n=5) to (n=2), (n=3) to (n=1) etc.,
• No two jumps will give a same line
• If there are a large number of atoms in a hydrogen sample, a large number of different jumps are possible
• Such a sample will give all the lines in all the series: Lyman, Balmer, Paschen etc.,
20. Brightness of the spectral lines:
• We will explain this using an example. We will write it in steps:
(i) Consider any one jump, say (n=4) to (n=2)
• This jump will emit a radiation of particular wavelength
(ii) Let the number of atoms in the hydrogen sample be large
• Then this particular jump will be taking place in so many atoms
(iii) As a result, we will get this particular wavelength from all those atoms
(iv) Then the line corresponding to that particular wavelength will appear bright
21. A very interesting information:
Consider the following two points (A) and (B):
(A) In each of the series (Lyman, Balmer Paschen etc.,), the right most line is far way from the rest of the lines. Also, the lines towards the left are closer to each other
(B) In the energy level diagram, ‘n=1’ is far below the rest. Also, the energy levels towards the top are closer to each other
• Is there any relation between (A) and (B)?
Let us check:
• We will check it by considering the Lyman series as an example. We will write it in steps:
(i) Five lines in the Lyman series are shown below:
• We know how to find the wave number (and hence the wavelengths) of each of those lines
• We simply use the expression given by Lyman: $\mathbf\small{\bar{\nu}=\frac{1}{\lambda}=109677\left(\frac{1}{1^2}-\frac{1}{n^2}\right)}$
(ii) Let us write the 'expression for wavelength' in terms of energy:
• We have:
Energy emitted/absorbed = $\mathbf\small{\Delta E=(E_f-E_i)=h\nu=\frac{hc}{\lambda}}$
$\mathbf\small{\Rightarrow \lambda=\frac{hc}{(E_f-E_i)}}$
• Using this expression, we can write the 'wavelength corresponding to any line' in any series
(iii) Consider the right most line in the Lyman series
• This is the line 1 in the above fig.2.33
• Bohr found out that, this line corresponds to the jump from (n=2) to (n=1). This is shown in fig.2.32
• Let us denote the 'wavelength corresponding to line 1' as $\mathbf\small{\lambda_1}$
• Then we can write: $\mathbf\small{\lambda_1=\frac{hc}{(E_1-E_2)}}$
(iv) Consider the second line from the right in the Lyman series
• This is the line 2 in the above fig.2.33
• Bohr found out that, this line corresponds to the jump from (n=3) to (n=1). This is shown in fig.2.32
• Let us denote the 'wavelength corresponding to line 2' as $\mathbf\small{\lambda_2}$
• Then we can write: $\mathbf\small{\lambda_2=\frac{hc}{(E_1-E_3)}}$
(v) In this way, we can make the following list:
$\mathbf\small{\lambda_1=\frac{hc}{(E_1-E_2)}}$
$\mathbf\small{\lambda_2=\frac{hc}{(E_1-E_3)}}$
$\mathbf\small{\lambda_3=\frac{hc}{(E_1-E_4)}}$
$\mathbf\small{\lambda_4=\frac{hc}{(E_1-E_5)}}$
$\mathbf\small{\lambda_5=\frac{hc}{(E_1-E_6)}}$
(vi) In the above list, we see a pattern:
• The numerators are all same: hc
• The denominators are: (E1-E2), (E1-E3), (E1-E4), (E1-E5) and (E1-E6)
• The first term is all the denominators are same: E1
(vii) We know that: E2 < E3 < E4 < E5 < E6
• That means., energies are progressively increasing
♦ But the increase is in the 'negative sense'
♦ For example: -12, -10, -7, -3, . . . is an increase in the 'negative sense'
(viii) So we get: (E1-E2) < (E1-E3) < (E1-E4) < (E1-E5) < (E1-E6)
• That means, the denominators are progressively increasing
• This increase is in the 'numerical value' only
• That is, this increase is effective when we ignore the sign
(ix) Since the denominators are increasing numerically, the net value decreases progressively
• That means, the wave lengths decrease progressively
■That is why, we are having to plot the lines from right towards the left in the '$\mathbf\small{\lambda}$ graph' shown in fig.2.33 above
• The calculations are shown in a tabular form below:
(x) Now consider the first two denominators (E1-E2) and (E1-E3)
• There is a large difference between E2 and E3
♦ This is visibly evident from the energy level diagram
• So there will be a large difference between (E1-E2) and (E1-E3)
• Consequently, there will be a large difference between $\mathbf\small{\lambda_1}$ and $\mathbf\small{\lambda_2}$
■So $\mathbf\small{\lambda_2}$ will be far away from $\mathbf\small{\lambda_1}$ in the '$\mathbf\small{\lambda}$ graph'
(xi) Now consider the last two denominators (E1-E5) and (E1-E6)
• There is not much difference between E5 and E6
♦ This is visibly evident from the energy level diagram
• So there will be not much difference between (E1-E5) and (E1-E6)
• Consequently, there will not be much difference between $\mathbf\small{\lambda_4}$ and $\mathbf\small{\lambda_5}$
■ So $\mathbf\small{\lambda_5}$ will be close to $\mathbf\small{\lambda_4}$ in the '$\mathbf\small{\lambda}$ graph'
■ Thus we see that, as we move towards the left in the '$\mathbf\small{\lambda}$ graph', the lines become closer and closer to each other
■ So the point 21(A) is indeed related to 21(B)
• We know that there are no atoms other than hydrogen, which have ‘only one electron’
• But there are ions. Let us see those ions in detail:
1. We know that helium (He) has 2 electrons
• If we remove one of those two electrons, the He atom will become a He+ ion
• The He+ ion has only one electron
2. We know that lithium (Li) has 3 electrons
• If we remove 2 of those 3 electrons, the Li atom will become a Li2+ ion
• The Li2+ ion has only one electron
3. We know that beryllium (Be) has 4 electrons
• If we remove 3 of those 4 electrons, the Be atom will become a Be3+ ion
• The Be3+ ion has only one electron
4. These types of ions described in (1), (2) and (3) are called hydrogen like species
• Bohr’s model is applicable to all hydrogen like species
5. We have seen that the expression for energy at various energy levels of the hydrogen atom: $\mathbf\small{E_n=-2.18\times 10^{-18}\left(\frac{1}{n^2}\right)\,\,\rm{J}}$
• By making a slight modification, this expression can be made available to all members of the hydrogen like species. Thus we get:
The general expression for energy at various ‘energy levels’ is: $\mathbf\small{E_n=-2.18\times 10^{-18}\left(\frac{Z^2}{n^2}\right)\,\,\rm{J}}$
6. We have seen that the expression for radii of various energy levels of the hydrogen atom: $\mathbf\small{r_n=52.9n^2\,\,\rm{pm}}$
• By making a slight modification, this expression can be made available to all members of the hydrogen like species. Thus we get:
The general expression for radii of various ‘energy levels; is: $\mathbf\small{r_n=\frac{52.9n^2}{Z}\,\,\rm{pm}}$
7. Consider the expression for energy that we wrote in (5). The Z is in the numerator
• So for the higher species:
♦ the atomic number increases
♦ and the result as a whole will become numerically larger
• That means, for higher species, the result as a whole will become more and more negative
• An example:
♦ E1 for hydrogen is given by: $\mathbf\small{E_1=-2.18\times 10^{-18}\left(\frac{1^2}{1^2}\right)}$ = -2.18 × 10-18 J
♦ E1 for helium is given by: $\mathbf\small{E_1=-2.18\times 10^{-18}\left(\frac{2^2}{1^2}\right)}$ = -8.72 × 10-18 J
• So in the energy level diagram, the E1 of helium will be above that of hydrogen
8. Consider the expression for radius that we wrote in (7). The Z is in the denominator
• So for higher species:
♦ the atomic number increases
♦ and the result as a whole will become smaller
• That means, for higher species, the radius will become smaller and smaller
• An example:
♦ r1 for hydrogen is given by: $\mathbf\small{r_n=\frac{52.9(1^2)}{1}}$ = 52.9 pm
♦ r1 for helium is given by: $\mathbf\small{r_n=\frac{52.9(1^2)}{2}}$ = 26.45 pm
• The reason for this type of decrease can be written as follows:
(i) Increase in Z means that, there are more protons inside the nucleus
(ii) So there will be an increased attraction on the electron
(iii) That means, it will be pulled towards the nucleus. This decreases the radius
9. We have seen the general expressions for energies and radii
• There is a general expression for velocity also. Using that expression, we can calculate the velocity with which the electron moves around the nucleus
• At this stage, we do not have to learn about that expression
• But it is worth to mention two points related to velocity:
♦ The velocity increases with Z
♦ The velocity decreases with n
Solved examples 2.41 to 2.47
Solved examples 2.48 to 2.53
Bohr model was a very good improvement over Rutherford’s model. But it had some limitations:
1. Consider the red line in the Balmer series
• When this red line is examined closely with high resolution devices, we see two closely spaced lines
• That means, the red line is in fact ‘two closely spaced lines’
• The gap between them will not be visible if we use devices of ordinary resolution
• Such a combination of two lines is called doublet in the atomic spectrum
• Using Bohr model, we cannot give an explanation for the formation of this doublet
2. Bohr model can explain the spectrum of hydrogen and hydrogen like species
• But it cannot explain the spectrum of atoms having more than one electron
3. In the presence of a magnetic field, the spectral lines split into different components
• This is called the Zeeman effect
• Bohr model cannot give an explanation for this phenomenon
4. In the presence of an electric field, the spectral lines split into different components
• This is called the Stark effect
• Bohr model cannot give an explanation for this phenomenon
5. A molecule is formed when two or more atoms combine
• Bohr model is not able to give a satisfactory explanation for how the combination between two or more atoms take place
1. In the postulate C, we saw this: $\mathbf\small{\Delta E = E_f-E_i}$
2. But postulate E gives the exact values of Ei and Ef:
• $\mathbf\small{E_f=-R_H \left(\frac{1}{n_f^2}\right)}$
• $\mathbf\small{E_i=-R_H \left(\frac{1}{n_i^2}\right)}$
3. Substituting these 'energies' in (1), we get:$\mathbf\small{\Delta E=\left[-R_H \left(\frac{1}{n_f^2}\right)--R_H \left(\frac{1}{n_i^2}\right)\right]}$
$\mathbf\small{\Rightarrow \Delta E=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
4. This expression in (3) gives the energy which is absorbed/emitted when the electron makes the jump. Let us apply it for absorption and emission separately:
■ If it is an absorption, the electron will be jumping from a lower state to a higher state
♦ In such a jump, nf > ni
♦ So the term inside the parenthesis will be positive
♦ Thus ΔE will be positive
■ If it is an emission, the electron will be jumping from a higher state to a lower state
♦ In such a jump, ni > nf
♦ So the term inside the parenthesis will be negative
♦ Thus ΔE will be negative
5. Thus we can write:
• A positive ΔE indicates absorption
• A negative ΔE indicates emission
6. Frequency of radiation:
• In the postulate C, we saw the Bohr’s frequency rule: $\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{(E_f-E_i)}{h}}$
• This rule gives the frequency of the radiation which is absorbed/emitted when the electron makes the jump
7. But postulate E gives the exact values of Ei and Ef:
• $\mathbf\small{E_f=-R_H \left(\frac{1}{n_f^2}\right)}$
• $\mathbf\small{E_i=-R_H \left(\frac{1}{n_i^2}\right)}$
8. Substituting these 'energies' in (6), we get:$\mathbf\small{\nu=\frac{\Delta E}{h}=\frac{1}{h}\left[-R_H \left(\frac{1}{n_f^2}\right)--R_H \left(\frac{1}{n_i^2}\right)\right]}$
$\mathbf\small{\Rightarrow \nu=\frac{R_H}{h}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
9. Substituting the values of RH and h in (8), we get:
$\mathbf\small{\nu=\frac{2.18\times 10^{-18}(J)}{6.626\times 10^{-34}(J s)}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
$\mathbf\small{\Rightarrow \nu=3.29\times 10^{15}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){s^{-1}}}$
• Thus we can calculate the frequency of the radiation absorbed/emitted when the electron makes the jump
10. Earlier, we saw the expression given by Rydberg: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)}$ (Details here)
• This expression is in terms of the 'wave number' $\mathbf\small{\bar{\nu}}$
♦ That is., the expression gives the wave number of the radiation absorbed/emitted
11. So let us rewrite our present expression also in terms of $\mathbf\small{\bar{\nu}}$• We have: $\mathbf\small{\nu=\frac{c}{\lambda}}$
$\mathbf\small{\Rightarrow \bar{\nu}=\frac{1}{\lambda}=\frac{\nu}{c}}$
• Thus our present equation in (9) becomes:
$\mathbf\small{\bar{\nu}=\frac{3.29\times 10^{15}}{3\times 10^{8}}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
$\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
■ This is the same equation derived by Rydberg
• However we must remember the two aspects:
(i) Bohr derived the equation by analyzing the internal structure of the hydrogen atom
(ii) Rydberg derived the equation with out any knowledge about the internal structure of the hydrogen atom. He used pure mathematics
12. Now we will see how Bohr’s model explains the lines seen in hydrogen spectrum
• We will use the lines in the Balmer series as an example
• We saw the Balmer series earlier in fig.2.23. (Details here) It is shown again below:
(i) This wave number was derived using the expression given by Balmer: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{2^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 3
(ii) Now consider the expression in (11). It is the equation derived by Bohr: $\mathbf\small{\bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
• If we put ni = 2 and nf = 3, we will get the same wave number
(We can easily convert cm-1 to m-1 and vice versa)
(iii) So according to Bohr, the radiation corresponding to the right most line in the Balmer series is emitted when:(We can easily convert cm-1 to m-1 and vice versa)
The electron jumps from the third energy level (n=3) to the second energy level (n=2)
14. We saw that, the 'second line from the right' represents a radiation of wave number 20564.4375 cm-1
(i) This wave number was derived using the expression given by Balmer: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{2^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 4
(ii) Now consider the expression derived by Bohr: $\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
• If we put ni = 2 and nf = 4, we will get the same wave number
(We can easily convert cm-1 to m-1 and vice versa)
(iii) So according to Bohr, the radiation corresponding to the second line from the right in the Balmer series is emitted when:(We can easily convert cm-1 to m-1 and vice versa)
The electron jumps from the fourth energy level (n=4) to the second energy level (n=2)
15. In this way, all lines in the Balmer series can be explained
• We will see one more example. This time, from another series, say Paschen series
• We saw that, the right most line in the Paschen series represents a radiation of wave number 1.87564 × 10-4 cm-1
(i) This wave number was derived using the expression given by Paschen: $\mathbf\small{\bar{\nu}=109677\left(\frac{1}{3^2}-\frac{1}{n^2}\right)}$
• For the right most line, we put n = 4
(ii) Now consider the expression derived by Bohr: $\mathbf\small{\Rightarrow \bar{\nu}=1.09677\times 10^{7}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right){m^{-1}}}$
• If we put ni = 3 and nf = 4, we will get the same wave number
(We can easily convert cm-1 to m-1 and vice versa)
(iii) So according to Bohr, the radiation corresponding to the right most line in the Paschen series is emitted when:(We can easily convert cm-1 to m-1 and vice versa)
The electron jumps from the fourth energy level (n=4) to the third energy level (n=3)
• In this way, all lines in the Paschen series can be explained
16. Thus we can write:
It was Neils Bohr who solved the mystery of 'how the lines are formed in the hydrogen spectrum'
17. We saw three examples:
(i) In (13), we saw this:
• Right most line in the Balmer series:
♦ The electron jumps from the third energy level (n=3) to the second energy level (n=2)
(ii) In (14), we saw this:
• Second line from the right in the Balmer series:
♦ The electron jumps from the fourth energy level (n=4) to the second energy level (n=2)
(iii) In (15), we saw this:
• Right most line in the Paschen series:
♦ The electron jumps from the third energy level (n=4) to the second energy level (n=3)
• We will mark these three 'jumps' in the energy level diagram that we saw in fig.2.29 in the previous section. The modified fig. is shown in fig.2.31 below:
 |
| Fig.2.31 |
• The jumps corresponding to Lyman, Blamer and Paschen series are shown in fig.2.32 below:
 |
| Fig.2.32 |
• Each line in the hydrogen spectrum (absorption spectrum as well as emission spectrum) can be related to 'a jump between two energy levels'
• Some of the jumps giving emission spectrum are:
(n=6) to (n=5), (n=5) to (n=2), (n=3) to (n=1) etc.,
• No two jumps will give a same line
• If there are a large number of atoms in a hydrogen sample, a large number of different jumps are possible
• Such a sample will give all the lines in all the series: Lyman, Balmer, Paschen etc.,
20. Brightness of the spectral lines:
• We will explain this using an example. We will write it in steps:
(i) Consider any one jump, say (n=4) to (n=2)
• This jump will emit a radiation of particular wavelength
(ii) Let the number of atoms in the hydrogen sample be large
• Then this particular jump will be taking place in so many atoms
(iii) As a result, we will get this particular wavelength from all those atoms
(iv) Then the line corresponding to that particular wavelength will appear bright
21. A very interesting information:
Consider the following two points (A) and (B):
(A) In each of the series (Lyman, Balmer Paschen etc.,), the right most line is far way from the rest of the lines. Also, the lines towards the left are closer to each other
(B) In the energy level diagram, ‘n=1’ is far below the rest. Also, the energy levels towards the top are closer to each other
• Is there any relation between (A) and (B)?
Let us check:
• We will check it by considering the Lyman series as an example. We will write it in steps:
(i) Five lines in the Lyman series are shown below:
 |
| Fig.2.33 |
• We simply use the expression given by Lyman: $\mathbf\small{\bar{\nu}=\frac{1}{\lambda}=109677\left(\frac{1}{1^2}-\frac{1}{n^2}\right)}$
(ii) Let us write the 'expression for wavelength' in terms of energy:
• We have:
Energy emitted/absorbed = $\mathbf\small{\Delta E=(E_f-E_i)=h\nu=\frac{hc}{\lambda}}$
$\mathbf\small{\Rightarrow \lambda=\frac{hc}{(E_f-E_i)}}$
• Using this expression, we can write the 'wavelength corresponding to any line' in any series
(iii) Consider the right most line in the Lyman series
• This is the line 1 in the above fig.2.33
• Bohr found out that, this line corresponds to the jump from (n=2) to (n=1). This is shown in fig.2.32
• Let us denote the 'wavelength corresponding to line 1' as $\mathbf\small{\lambda_1}$
• Then we can write: $\mathbf\small{\lambda_1=\frac{hc}{(E_1-E_2)}}$
(iv) Consider the second line from the right in the Lyman series
• This is the line 2 in the above fig.2.33
• Bohr found out that, this line corresponds to the jump from (n=3) to (n=1). This is shown in fig.2.32
• Let us denote the 'wavelength corresponding to line 2' as $\mathbf\small{\lambda_2}$
• Then we can write: $\mathbf\small{\lambda_2=\frac{hc}{(E_1-E_3)}}$
(v) In this way, we can make the following list:
$\mathbf\small{\lambda_1=\frac{hc}{(E_1-E_2)}}$
$\mathbf\small{\lambda_2=\frac{hc}{(E_1-E_3)}}$
$\mathbf\small{\lambda_3=\frac{hc}{(E_1-E_4)}}$
$\mathbf\small{\lambda_4=\frac{hc}{(E_1-E_5)}}$
$\mathbf\small{\lambda_5=\frac{hc}{(E_1-E_6)}}$
(vi) In the above list, we see a pattern:
• The numerators are all same: hc
• The denominators are: (E1-E2), (E1-E3), (E1-E4), (E1-E5) and (E1-E6)
• The first term is all the denominators are same: E1
(vii) We know that: E2 < E3 < E4 < E5 < E6
• That means., energies are progressively increasing
♦ But the increase is in the 'negative sense'
♦ For example: -12, -10, -7, -3, . . . is an increase in the 'negative sense'
(viii) So we get: (E1-E2) < (E1-E3) < (E1-E4) < (E1-E5) < (E1-E6)
• That means, the denominators are progressively increasing
• This increase is in the 'numerical value' only
• That is, this increase is effective when we ignore the sign
(ix) Since the denominators are increasing numerically, the net value decreases progressively
• That means, the wave lengths decrease progressively
■That is why, we are having to plot the lines from right towards the left in the '$\mathbf\small{\lambda}$ graph' shown in fig.2.33 above
• The calculations are shown in a tabular form below:
• There is a large difference between E2 and E3
♦ This is visibly evident from the energy level diagram
• So there will be a large difference between (E1-E2) and (E1-E3)
• Consequently, there will be a large difference between $\mathbf\small{\lambda_1}$ and $\mathbf\small{\lambda_2}$
■So $\mathbf\small{\lambda_2}$ will be far away from $\mathbf\small{\lambda_1}$ in the '$\mathbf\small{\lambda}$ graph'
(xi) Now consider the last two denominators (E1-E5) and (E1-E6)
• There is not much difference between E5 and E6
♦ This is visibly evident from the energy level diagram
• So there will be not much difference between (E1-E5) and (E1-E6)
• Consequently, there will not be much difference between $\mathbf\small{\lambda_4}$ and $\mathbf\small{\lambda_5}$
■ So $\mathbf\small{\lambda_5}$ will be close to $\mathbf\small{\lambda_4}$ in the '$\mathbf\small{\lambda}$ graph'
■ Thus we see that, as we move towards the left in the '$\mathbf\small{\lambda}$ graph', the lines become closer and closer to each other
■ So the point 21(A) is indeed related to 21(B)
• So we now know how Neils Bohr effectively explained the structure of hydrogen atom
• The hydrogen atom has only one electron
■ Bohr’s model can be applied to all cases where there is ‘only one electron’
• The hydrogen atom has only one electron
■ Bohr’s model can be applied to all cases where there is ‘only one electron’
• But there are ions. Let us see those ions in detail:
1. We know that helium (He) has 2 electrons
• If we remove one of those two electrons, the He atom will become a He+ ion
• The He+ ion has only one electron
2. We know that lithium (Li) has 3 electrons
• If we remove 2 of those 3 electrons, the Li atom will become a Li2+ ion
• The Li2+ ion has only one electron
3. We know that beryllium (Be) has 4 electrons
• If we remove 3 of those 4 electrons, the Be atom will become a Be3+ ion
• The Be3+ ion has only one electron
4. These types of ions described in (1), (2) and (3) are called hydrogen like species
• Bohr’s model is applicable to all hydrogen like species
5. We have seen that the expression for energy at various energy levels of the hydrogen atom: $\mathbf\small{E_n=-2.18\times 10^{-18}\left(\frac{1}{n^2}\right)\,\,\rm{J}}$
• By making a slight modification, this expression can be made available to all members of the hydrogen like species. Thus we get:
The general expression for energy at various ‘energy levels’ is: $\mathbf\small{E_n=-2.18\times 10^{-18}\left(\frac{Z^2}{n^2}\right)\,\,\rm{J}}$
6. We have seen that the expression for radii of various energy levels of the hydrogen atom: $\mathbf\small{r_n=52.9n^2\,\,\rm{pm}}$
• By making a slight modification, this expression can be made available to all members of the hydrogen like species. Thus we get:
The general expression for radii of various ‘energy levels; is: $\mathbf\small{r_n=\frac{52.9n^2}{Z}\,\,\rm{pm}}$
7. Consider the expression for energy that we wrote in (5). The Z is in the numerator
• So for the higher species:
♦ the atomic number increases
♦ and the result as a whole will become numerically larger
• That means, for higher species, the result as a whole will become more and more negative
• An example:
♦ E1 for hydrogen is given by: $\mathbf\small{E_1=-2.18\times 10^{-18}\left(\frac{1^2}{1^2}\right)}$ = -2.18 × 10-18 J
♦ E1 for helium is given by: $\mathbf\small{E_1=-2.18\times 10^{-18}\left(\frac{2^2}{1^2}\right)}$ = -8.72 × 10-18 J
• So in the energy level diagram, the E1 of helium will be above that of hydrogen
8. Consider the expression for radius that we wrote in (7). The Z is in the denominator
• So for higher species:
♦ the atomic number increases
♦ and the result as a whole will become smaller
• That means, for higher species, the radius will become smaller and smaller
• An example:
♦ r1 for hydrogen is given by: $\mathbf\small{r_n=\frac{52.9(1^2)}{1}}$ = 52.9 pm
♦ r1 for helium is given by: $\mathbf\small{r_n=\frac{52.9(1^2)}{2}}$ = 26.45 pm
• The reason for this type of decrease can be written as follows:
(i) Increase in Z means that, there are more protons inside the nucleus
(ii) So there will be an increased attraction on the electron
(iii) That means, it will be pulled towards the nucleus. This decreases the radius
9. We have seen the general expressions for energies and radii
• There is a general expression for velocity also. Using that expression, we can calculate the velocity with which the electron moves around the nucleus
• At this stage, we do not have to learn about that expression
• But it is worth to mention two points related to velocity:
♦ The velocity increases with Z
♦ The velocity decreases with n
The links given below can be used to see some solved examples presented in pdf format
Solved examples 2.41 to 2.47
Solved examples 2.48 to 2.53
Limitations of Bohr's model
1. Consider the red line in the Balmer series
• When this red line is examined closely with high resolution devices, we see two closely spaced lines
• That means, the red line is in fact ‘two closely spaced lines’
• The gap between them will not be visible if we use devices of ordinary resolution
• Such a combination of two lines is called doublet in the atomic spectrum
• Using Bohr model, we cannot give an explanation for the formation of this doublet
2. Bohr model can explain the spectrum of hydrogen and hydrogen like species
• But it cannot explain the spectrum of atoms having more than one electron
3. In the presence of a magnetic field, the spectral lines split into different components
• This is called the Zeeman effect
• Bohr model cannot give an explanation for this phenomenon
4. In the presence of an electric field, the spectral lines split into different components
• This is called the Stark effect
• Bohr model cannot give an explanation for this phenomenon
5. A molecule is formed when two or more atoms combine
• Bohr model is not able to give a satisfactory explanation for how the combination between two or more atoms take place
This completes our present discussion on the 'basics of Bohr model of hydrogen atom' and 'hydrogen spectrum'. We also saw the limitations of Bohr model. In the next section, we will see some developments that enabled scientists to arrive at a satisfactory model
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