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When an electron in a hydrogen atom transitions from a higher energy level ([tex]\( n_{initial} \)[/tex]) to a lower energy level ([tex]\( n_{final} \)[/tex]), it emits light of a specific wavelength. The wavelength ([tex]\( \lambda \)[/tex]) can be calculated using the Rydberg formula:
[tex]\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \][/tex]
where:
- [tex]\( R \)[/tex] is the Rydberg constant ([tex]\( 1.097373 \times 10^7 \, \text{m}^{-1} \)[/tex])
- [tex]\( n_{initial} \)[/tex] is the initial energy level of the electron (5 in this case)
- [tex]\( n_{final} \)[/tex] is the final energy level of the electron (1 in this case)
1. Calculate the change in energy levels:
[tex]\[ \frac{1}{5^2} = \frac{1}{25} \][/tex]
[tex]\[ \frac{1}{1^2} = 1 \][/tex]
So the change in terms of squared energy levels is:
[tex]\[ \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} = 1 - \frac{1}{25} \][/tex]
[tex]\[ = 1 - 0.04 \][/tex]
[tex]\[ = 0.96 \][/tex]
2. Using the Rydberg formula, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = 1.097373 \times 10^7 \times 0.96 \][/tex]
[tex]\[ = 1.05347728 \times 10^7 \, \text{m}^{-1} \][/tex]
3. Determine the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{1}{1.05347728 \times 10^7} \][/tex]
[tex]\[ \lambda \approx 9.492366466704273 \times 10^{-8} \, \text{meters} \][/tex]
[tex]\[ \lambda \approx 94.92 \, \text{nm} \][/tex] (Note: [tex]\( 1 \, nm = 10^{-9} \, m \)[/tex])
4. Calculate the frequency ([tex]\( f \)[/tex]) of the light emitted:
Frequency ([tex]\( f \)[/tex]) can be found using the relationship between the speed of light [tex]\( c \)[/tex] and the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ c = \lambda f \][/tex]
where:
- [tex]\( c \)[/tex] is the speed of light ([tex]\( 3.0 \times 10^8 \, \text{m/s} \)[/tex])
- [tex]\( \lambda \)[/tex] is the wavelength (from the previous step)
Rearranging to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
[tex]\[ = \frac{3.0 \times 10^8}{9.492366466704273 \times 10^{-8}} \][/tex]
[tex]\[ \approx 3.16043424 \times 10^{15} \, \text{Hz} \][/tex]
Therefore, the wavelength of the light emitted during the electron transition from [tex]\( n=5 \)[/tex] to [tex]\( n=1 \)[/tex] is approximately [tex]\( 9.492366466704273 \times 10^{-8} \)[/tex] meters (or 94.92 nm), and the frequency of the light is approximately [tex]\( 3.16043424 \times 10^{15} \)[/tex] Hz.
When an electron in a hydrogen atom transitions from a higher energy level ([tex]\( n_{initial} \)[/tex]) to a lower energy level ([tex]\( n_{final} \)[/tex]), it emits light of a specific wavelength. The wavelength ([tex]\( \lambda \)[/tex]) can be calculated using the Rydberg formula:
[tex]\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \][/tex]
where:
- [tex]\( R \)[/tex] is the Rydberg constant ([tex]\( 1.097373 \times 10^7 \, \text{m}^{-1} \)[/tex])
- [tex]\( n_{initial} \)[/tex] is the initial energy level of the electron (5 in this case)
- [tex]\( n_{final} \)[/tex] is the final energy level of the electron (1 in this case)
1. Calculate the change in energy levels:
[tex]\[ \frac{1}{5^2} = \frac{1}{25} \][/tex]
[tex]\[ \frac{1}{1^2} = 1 \][/tex]
So the change in terms of squared energy levels is:
[tex]\[ \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} = 1 - \frac{1}{25} \][/tex]
[tex]\[ = 1 - 0.04 \][/tex]
[tex]\[ = 0.96 \][/tex]
2. Using the Rydberg formula, calculate [tex]\( \frac{1}{\lambda} \)[/tex]:
[tex]\[ \frac{1}{\lambda} = 1.097373 \times 10^7 \times 0.96 \][/tex]
[tex]\[ = 1.05347728 \times 10^7 \, \text{m}^{-1} \][/tex]
3. Determine the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{1}{1.05347728 \times 10^7} \][/tex]
[tex]\[ \lambda \approx 9.492366466704273 \times 10^{-8} \, \text{meters} \][/tex]
[tex]\[ \lambda \approx 94.92 \, \text{nm} \][/tex] (Note: [tex]\( 1 \, nm = 10^{-9} \, m \)[/tex])
4. Calculate the frequency ([tex]\( f \)[/tex]) of the light emitted:
Frequency ([tex]\( f \)[/tex]) can be found using the relationship between the speed of light [tex]\( c \)[/tex] and the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ c = \lambda f \][/tex]
where:
- [tex]\( c \)[/tex] is the speed of light ([tex]\( 3.0 \times 10^8 \, \text{m/s} \)[/tex])
- [tex]\( \lambda \)[/tex] is the wavelength (from the previous step)
Rearranging to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
[tex]\[ = \frac{3.0 \times 10^8}{9.492366466704273 \times 10^{-8}} \][/tex]
[tex]\[ \approx 3.16043424 \times 10^{15} \, \text{Hz} \][/tex]
Therefore, the wavelength of the light emitted during the electron transition from [tex]\( n=5 \)[/tex] to [tex]\( n=1 \)[/tex] is approximately [tex]\( 9.492366466704273 \times 10^{-8} \)[/tex] meters (or 94.92 nm), and the frequency of the light is approximately [tex]\( 3.16043424 \times 10^{15} \)[/tex] Hz.
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