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A Rydberg atom is a single-electron atom...

A Rydberg atom is a single-electron atom with a large quantum number n. Rydberg states are close together in energy, so transitions between from a state `n + 1` to a state n in hydrogen. The wavelength of the emitted photon varies with quantum number n as

A

`(1)/(n^(2))`

B

`n`

C

`(1)/(n^(3))`

D

`n^(3)`

Text Solution

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The correct Answer is:
To solve the problem regarding the wavelength of the emitted photon during the transition from a Rydberg state \( n+1 \) to \( n \) in hydrogen, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Energy Levels**: The energy of an electron in a hydrogen atom at a principal quantum number \( n \) is given by: \[ E_n = -\frac{E_0}{n^2} \] where \( E_0 \) is a constant (approximately \( 13.6 \, \text{eV} \)). 2. **Calculate the Energy Difference**: The energy difference between the states \( n+1 \) and \( n \) is: \[ E_{n+1} - E_n = \left(-\frac{E_0}{(n+1)^2}\right) - \left(-\frac{E_0}{n^2}\right) \] This simplifies to: \[ E_{n+1} - E_n = E_0 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) \] 3. **Use the Formula for Energy Difference**: The expression for the energy difference can be rewritten using the formula for the difference of squares: \[ \frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{2n + 1}{n^2(n+1)^2} \] Thus, we have: \[ E_{n+1} - E_n = E_0 \cdot \frac{2n + 1}{n^2(n+1)^2} \] 4. **Relate Energy to Wavelength**: The energy of the emitted photon is also given by: \[ E = \frac{hc}{\lambda} \] Setting the two expressions for energy equal gives: \[ \frac{hc}{\lambda} = E_0 \cdot \frac{2n + 1}{n^2(n+1)^2} \] 5. **Solve for Wavelength**: Rearranging for \( \lambda \): \[ \lambda = \frac{hc \cdot n^2(n+1)^2}{E_0(2n + 1)} \] 6. **Approximate for Large n**: For large values of \( n \), \( n+1 \approx n \) and \( 2n + 1 \approx 2n \). Thus, we can simplify: \[ \lambda \approx \frac{hc \cdot n^2 n^2}{E_0 \cdot 2n} = \frac{hc \cdot n^4}{2E_0} \] Therefore, we can conclude that: \[ \lambda \propto n^3 \] ### Final Result: The wavelength of the emitted photon varies with the quantum number \( n \) as: \[ \lambda \propto n^3 \]

To solve the problem regarding the wavelength of the emitted photon during the transition from a Rydberg state \( n+1 \) to \( n \) in hydrogen, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Energy Levels**: The energy of an electron in a hydrogen atom at a principal quantum number \( n \) is given by: \[ E_n = -\frac{E_0}{n^2} ...
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