Calculate the energy released in H-alpha emission of hydrogen.
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![**How to Calculate the Energy Released in H-alpha Emission of Hydrogen**
When an electron in a hydrogen atom transitions from a higher energy level (n) to a lower energy level, it releases energy in the form of light. Specifically, the H-alpha emission occurs when the electron falls from the n=3 energy level to the n=2 energy level. This process can be explained using the principles of quantum mechanics and the Rydberg formula.
To calculate the energy released during the H-alpha emission of hydrogen, follow these steps:
1. **Understand the Rydberg Formula**:
\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
where:
- \( \lambda \) is the wavelength of the emitted light,
- \( R \) is the Rydberg constant (1.097 x 10^7 m^-1),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's initial and final energy levels (with \( n_2 > n_1 \)).
2. **Substitute the Quantum Numbers for H-alpha**:
For H-alpha emission, \( n_1 = 2 \) and \( n_2 = 3 \).
\[
\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{5}{36} \right)
\]
3. **Calculate the Wavelength (λ)**:
\[
\lambda = \frac{36}{5R}
\]
4. **Determine the Energy Released (E)**:
The energy of the emitted photon can be calculated using the equation:
\[
E = \frac{hc}{\lambda}
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Transcribed Image Text:**How to Calculate the Energy Released in H-alpha Emission of Hydrogen**
When an electron in a hydrogen atom transitions from a higher energy level (n) to a lower energy level, it releases energy in the form of light. Specifically, the H-alpha emission occurs when the electron falls from the n=3 energy level to the n=2 energy level. This process can be explained using the principles of quantum mechanics and the Rydberg formula.
To calculate the energy released during the H-alpha emission of hydrogen, follow these steps:
1. **Understand the Rydberg Formula**:
\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
where:
- \( \lambda \) is the wavelength of the emitted light,
- \( R \) is the Rydberg constant (1.097 x 10^7 m^-1),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's initial and final energy levels (with \( n_2 > n_1 \)).
2. **Substitute the Quantum Numbers for H-alpha**:
For H-alpha emission, \( n_1 = 2 \) and \( n_2 = 3 \).
\[
\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right)
\]
\[
\frac{1}{\lambda} = R \left( \frac{5}{36} \right)
\]
3. **Calculate the Wavelength (λ)**:
\[
\lambda = \frac{36}{5R}
\]
4. **Determine the Energy Released (E)**:
The energy of the emitted photon can be calculated using the equation:
\[
E = \frac{hc}{\lambda}
\
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