For a hydrogen atom, calculate the energy of the photon that would be emitted for the orbital transition of n(initial) = 5 to n(final) = 2. The Rydberg constant is 1.09678 x 10² m-¹.

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**Question 1 of 21**

For a hydrogen atom, calculate the energy of the photon that would be emitted for the orbital transition of n(initial) = 5 to n(final) = 2. The Rydberg constant is \(1.09678 \times 10^7 \, \text{m}^{-1}\).

**Explanation:**

This question involves calculating the energy of a photon emitted when an electron transitions between energy levels in a hydrogen atom. Specifically, it asks for the energy when an electron drops from the 5th energy level (n=5) to the 2nd energy level (n=2). The Rydberg constant provided is crucial for this calculation.

**Calculation:**
To find the energy (E) of the emitted photon, we use the Rydberg formula for hydrogen:

\[
E = h \cdot c \cdot \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \cdot R
\]

Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\))
- \(R\) is the Rydberg constant (\(1.09678 \times 10^7 \, \text{m}^{-1}\))
- \(n_{\text{final}}\) and \(n_{\text{initial}}\) are the final and initial energy levels, respectively.

The keypad shown illustrates where the answer is to be entered, with a field labeled "J" indicating the unit for energy (Joules).
Transcribed Image Text:**Question 1 of 21** For a hydrogen atom, calculate the energy of the photon that would be emitted for the orbital transition of n(initial) = 5 to n(final) = 2. The Rydberg constant is \(1.09678 \times 10^7 \, \text{m}^{-1}\). **Explanation:** This question involves calculating the energy of a photon emitted when an electron transitions between energy levels in a hydrogen atom. Specifically, it asks for the energy when an electron drops from the 5th energy level (n=5) to the 2nd energy level (n=2). The Rydberg constant provided is crucial for this calculation. **Calculation:** To find the energy (E) of the emitted photon, we use the Rydberg formula for hydrogen: \[ E = h \cdot c \cdot \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \cdot R \] Where: - \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) - \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)) - \(R\) is the Rydberg constant (\(1.09678 \times 10^7 \, \text{m}^{-1}\)) - \(n_{\text{final}}\) and \(n_{\text{initial}}\) are the final and initial energy levels, respectively. The keypad shown illustrates where the answer is to be entered, with a field labeled "J" indicating the unit for energy (Joules).
Expert Solution
Step 1: Hydrogen spectrum

The electronic energy levels in an atom are quantized and when an electron undergoes a transition from a higher energy level to a lower energy level a photon of electromagnetic radiation is emitted. The series of the radiation in the emission spectrum is known as the hydrogen emission spectrum.

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