Calculate the wavelength frequency and energy on transition from n =6 in the paschen series

In the **Paschen series** of the hydrogen atom, transitions occur when electrons move to the \( n = 3 \) energy level from higher energy levels. For a transition from \( n = 6 \) to \( n = 3 \), we can calculate the **wavelength**, **frequency**, and **energy** using the following steps:

### 1. Wavelength Calculation (using the Rydberg formula):
The wavelength \( \lambda \) for a transition between two energy levels in hydrogen can be calculated using the Rydberg formula:

\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)
\]

Where:
- \( R_H \) is the Rydberg constant \( = 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 = 6 \) (initial energy level)
- \( n_2 = 3 \) (final energy level)

Let's calculate \( \lambda \):

\[
\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{6^2} \right)
\]

### 2. Frequency Calculation:
Once we have the wavelength, the frequency \( \nu \) can be found using the speed of light equation:

\[
\nu = \frac{c}{\lambda}
\]

Where:
- \( c = 3.00 \times 10^8 \, \text{m/s} \) (speed of light)

### 3. Energy Calculation:
The energy \( E \) of the photon can be calculated using the Planck's equation:

\[
E = h \nu
\]

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J·s} \) (Planck’s constant)

Now, let's perform the calculations.

#### Step 1: Calculate Wavelength \( \lambda \) 
The wavelength \( \lambda \) for the transition from \( n = 6 \) to \( n = 3 \) in the Paschen series is approximately \( 1.09 \times 10^{-6} \, \text{m} \) (or 1093.89 nm), which lies in the infrared region of the electromagnetic spectrum.

Now let's calculate the frequency \( \nu \) and the energy \( E \) of the photon.

#### Step 2: Calculate Frequency \( \nu \)The frequency \( \nu \) of the photon for this transition is approximately \( 2.74 \times 10^{14} \, \text{Hz} \).

#### Step 3: Calculate Energy \( E \)The energy \( E \) of the photon for the transition from \( n = 6 \) to \( n = 3 \) is approximately \( 1.82 \times 10^{-19} \, \text{J} \).

### Summary:
- **Wavelength**: \( 1.09 \times 10^{-6} \, \text{m} \) (1093.89 nm)
- **Frequency**: \( 2.74 \times 10^{14} \, \text{Hz} \)
- **Energy**: \( 1.82 \times 10^{-19} \, \text{J} \)

Would you like more details on any part of this? Here are some related questions:

1. How does the energy of a photon relate to its wavelength?
2. Can the Paschen series transitions be observed in visible light?
3. What is the difference between the Paschen and Balmer series?
4. How is the Rydberg constant derived?
5. Why do energy levels get closer as \( n \) increases?

**Tip**: The shorter the wavelength, the higher the energy of the photon.

Calculate the wavelength, frequency, and energy on transition from n = 6 in the Paschen series.

Learn how to calculate the wavelength, frequency, and energy for the transition from n=6 to n=3 in the Paschen series of the hydrogen atom. This detailed solution uses the Rydberg formula, the speed of light equation, and Planck's equation to explain key concepts in quantum mechanics. Suitable for students in Grades 11-12.

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