Answer :
Let's use Bohr's equation to calculate the wavelength of light emitted in each of the given electronic transitions. The equation is:
[tex]\[ \frac{1}{\lambda} = R \left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] \][/tex]
where:
- [tex]\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex]
- [tex]\( n_1 \)[/tex] is the initial energy level
- [tex]\( n_2 \)[/tex] is the final energy level
- [tex]\( \lambda \)[/tex] is the wavelength of the emitted light
### i. Transition: [tex]\( n=2 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 2 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
First, calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{3}{4} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{3}{4} = 8.2275 \times 10^6 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{8.2275 \times 10^6 \, \text{m}^{-1}} \approx 1.215436037678517 \times 10^{-7} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 1.215436037678517 \times 10^{-7} \, \text{m} \times 10^9 \approx 121.5436037678517 \, \text{nm} \][/tex]
### ii. Transition: [tex]\( n=4 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 4 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
Calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{4^2} = 1 - \frac{1}{16} = \frac{15}{16} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{15}{16} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{15}{16} = 1.0284375 \times 10^7 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{1.0284375 \times 10^7 \, \text{m}^{-1}} \approx 9.723488301428137 \times 10^{-8} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 9.723488301428137 \times 10^{-8} \, \text{m} \times 10^9 \approx 97.23488301428137 \, \text{nm} \][/tex]
### iii. Transition: [tex]\( n=6 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 6 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
Calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{6^2} = 1 - \frac{1}{36} = \frac{35}{36} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{35}{36} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{35}{36} = 1.067986111 \times 10^7 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{1.067986111 \times 10^7 \, \text{m}^{-1}} \approx 9.376220862091418 \times 10^{-8} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 9.376220862091418 \times 10^{-8} \, \text{m} \times 10^9 \approx 93.76220862091418 \, \text{nm} \][/tex]
### Summary
i. [tex]\( n=2 \rightarrow n=1 \)[/tex]: approximately [tex]\( 121.5436 \, \text{nm} \)[/tex]
ii. [tex]\( n=4 \rightarrow n=1 \)[/tex]: approximately [tex]\( 97.23488 \, \text{nm} \)[/tex]
iii. [tex]\( n=6 \rightarrow n=1 \)[/tex]: approximately [tex]\( 93.76221 \, \text{nm} \)[/tex]
[tex]\[ \frac{1}{\lambda} = R \left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] \][/tex]
where:
- [tex]\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex]
- [tex]\( n_1 \)[/tex] is the initial energy level
- [tex]\( n_2 \)[/tex] is the final energy level
- [tex]\( \lambda \)[/tex] is the wavelength of the emitted light
### i. Transition: [tex]\( n=2 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 2 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
First, calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{3}{4} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{3}{4} = 8.2275 \times 10^6 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{8.2275 \times 10^6 \, \text{m}^{-1}} \approx 1.215436037678517 \times 10^{-7} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 1.215436037678517 \times 10^{-7} \, \text{m} \times 10^9 \approx 121.5436037678517 \, \text{nm} \][/tex]
### ii. Transition: [tex]\( n=4 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 4 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
Calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{4^2} = 1 - \frac{1}{16} = \frac{15}{16} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{15}{16} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{15}{16} = 1.0284375 \times 10^7 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{1.0284375 \times 10^7 \, \text{m}^{-1}} \approx 9.723488301428137 \times 10^{-8} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 9.723488301428137 \times 10^{-8} \, \text{m} \times 10^9 \approx 97.23488301428137 \, \text{nm} \][/tex]
### iii. Transition: [tex]\( n=6 \rightarrow n=1 \)[/tex]
For this transition:
- [tex]\( n_1 = 6 \)[/tex]
- [tex]\( n_2 = 1 \)[/tex]
Calculate the right-hand side of the equation inside the brackets:
[tex]\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = \frac{1}{1^2} - \frac{1}{6^2} = 1 - \frac{1}{36} = \frac{35}{36} \][/tex]
Now, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = R \left[ \frac{35}{36} \right] = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{35}{36} = 1.067986111 \times 10^7 \, \text{m}^{-1} \][/tex]
Therefore, the wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ \lambda = \frac{1}{1.067986111 \times 10^7 \, \text{m}^{-1}} \approx 9.376220862091418 \times 10^{-8} \, \text{m} \][/tex]
Convert this wavelength to nanometers:
[tex]\[ \lambda_{nm} = \lambda_{m} \times 10^9 = 9.376220862091418 \times 10^{-8} \, \text{m} \times 10^9 \approx 93.76220862091418 \, \text{nm} \][/tex]
### Summary
i. [tex]\( n=2 \rightarrow n=1 \)[/tex]: approximately [tex]\( 121.5436 \, \text{nm} \)[/tex]
ii. [tex]\( n=4 \rightarrow n=1 \)[/tex]: approximately [tex]\( 97.23488 \, \text{nm} \)[/tex]
iii. [tex]\( n=6 \rightarrow n=1 \)[/tex]: approximately [tex]\( 93.76221 \, \text{nm} \)[/tex]
