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To calculate the frequency of the light emitted when an electron in a hydrogen atom transitions from [tex]\( n=4 \)[/tex] to [tex]\( n=3 \)[/tex], follow these steps:
1. Identify the constants involved:
- Planck constant, [tex]\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]
- Speed of light, [tex]\( c = 3.0 \times 10^8 \, \text{m/s} \)[/tex]
- Rydberg constant, [tex]\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex]
2. Transition levels:
- Initial level, [tex]\( n_1 = 4 \)[/tex]
- Final level, [tex]\( n_2 = 3 \)[/tex]
3. Use the Rydberg formula to calculate the energy difference:
The energy difference [tex]\( \Delta E \)[/tex] between two levels in a hydrogen atom is given by:
[tex]\[ \Delta E = R h c \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \][/tex]
Plugging in the values:
[tex]\[ \Delta E = (1.097 \times 10^7 \, \text{m}^{-1}) \times (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.0 \times 10^8 \, \text{m/s}) \times \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \][/tex]
4. Calculate the difference in fractions:
[tex]\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} \][/tex]
[tex]\[ \frac{1}{9} = 0.1111 \quad \text{and} \quad \frac{1}{16} = 0.0625 \][/tex]
[tex]\[ 0.1111 - 0.0625 = 0.0486 \][/tex]
5. Continue to compute [tex]\(\Delta E\)[/tex]:
[tex]\[ \Delta E = (1.097 \times 10^7) \times (6.626 \times 10^{-34}) \times (3.0 \times 10^8) \times 0.0486 \][/tex]
[tex]\[ \Delta E \approx 1.060 \times 10^{-19} \, \text{J} \][/tex]
6. Calculate the frequency of the emitted light:
The frequency ([tex]\( \nu \)[/tex]) of the emitted light can be found using the energy relationship:
[tex]\[ \Delta E = h \nu \][/tex]
Solving for [tex]\( \nu \)[/tex]:
[tex]\[ \nu = \frac{\Delta E}{h} \][/tex]
Substituting the values:
[tex]\[ \nu = \frac{1.060 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \][/tex]
[tex]\[ \nu \approx 1.600 \times 10^{14} \, \text{Hz} \][/tex]
[tex]\[ \nu \approx 1.5998 \times 10^{14} \, \text{Hz} \][/tex]
Therefore, the frequency of the emitted light when the electron transitions from [tex]\( n=4 \)[/tex] to [tex]\( n=3 \)[/tex] in a hydrogen atom is approximately [tex]\( 1.5998 \times 10^{14} \, \text{Hz} \)[/tex]. The energy difference corresponding to this transition is approximately [tex]\( 1.060 \times 10^{-19} \, \text{J} \)[/tex].
1. Identify the constants involved:
- Planck constant, [tex]\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]
- Speed of light, [tex]\( c = 3.0 \times 10^8 \, \text{m/s} \)[/tex]
- Rydberg constant, [tex]\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex]
2. Transition levels:
- Initial level, [tex]\( n_1 = 4 \)[/tex]
- Final level, [tex]\( n_2 = 3 \)[/tex]
3. Use the Rydberg formula to calculate the energy difference:
The energy difference [tex]\( \Delta E \)[/tex] between two levels in a hydrogen atom is given by:
[tex]\[ \Delta E = R h c \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \][/tex]
Plugging in the values:
[tex]\[ \Delta E = (1.097 \times 10^7 \, \text{m}^{-1}) \times (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.0 \times 10^8 \, \text{m/s}) \times \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \][/tex]
4. Calculate the difference in fractions:
[tex]\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} \][/tex]
[tex]\[ \frac{1}{9} = 0.1111 \quad \text{and} \quad \frac{1}{16} = 0.0625 \][/tex]
[tex]\[ 0.1111 - 0.0625 = 0.0486 \][/tex]
5. Continue to compute [tex]\(\Delta E\)[/tex]:
[tex]\[ \Delta E = (1.097 \times 10^7) \times (6.626 \times 10^{-34}) \times (3.0 \times 10^8) \times 0.0486 \][/tex]
[tex]\[ \Delta E \approx 1.060 \times 10^{-19} \, \text{J} \][/tex]
6. Calculate the frequency of the emitted light:
The frequency ([tex]\( \nu \)[/tex]) of the emitted light can be found using the energy relationship:
[tex]\[ \Delta E = h \nu \][/tex]
Solving for [tex]\( \nu \)[/tex]:
[tex]\[ \nu = \frac{\Delta E}{h} \][/tex]
Substituting the values:
[tex]\[ \nu = \frac{1.060 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \][/tex]
[tex]\[ \nu \approx 1.600 \times 10^{14} \, \text{Hz} \][/tex]
[tex]\[ \nu \approx 1.5998 \times 10^{14} \, \text{Hz} \][/tex]
Therefore, the frequency of the emitted light when the electron transitions from [tex]\( n=4 \)[/tex] to [tex]\( n=3 \)[/tex] in a hydrogen atom is approximately [tex]\( 1.5998 \times 10^{14} \, \text{Hz} \)[/tex]. The energy difference corresponding to this transition is approximately [tex]\( 1.060 \times 10^{-19} \, \text{J} \)[/tex].
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