What is the energy difference and the frequency of light emitted when the electron in a hydrogen atom undergoes transition from the energy level `n = 4` to the energy `n = 3` given that the value of Rydberg constant is `1.0974 xx 10^(7)m^(-1)`?

Explanation

To find the energy difference and the frequency of light emitted when an electron in a hydrogen atom transitions from the energy level \( n = 4 \) to \( n = 3 \), we can follow these steps:

### Step 1: Use the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during a transition in a hydrogen atom is given by:

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

Where:
- \( R \) is the Rydberg constant \( (1.0974 \times 10^7 \, \text{m}^{-1}) \)
- \( n_1 = 3 \) (final energy level)
- \( n_2 = 4 \) (initial energy level)

### Step 2: Calculate the Wavelength
Substituting the values into the formula:

\[ \frac{1}{\lambda} = 1.0974 \times 10^7 \cdot \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

Calculating \( \frac{1}{3^2} \) and \( \frac{1}{4^2} \):

\[ \frac{1}{3^2} = \frac{1}{9} \quad \text{and} \quad \frac{1}{4^2} = \frac{1}{16} \]

Finding a common denominator (144):

\[ \frac{1}{9} = \frac{16}{144} \quad \text{and} \quad \frac{1}{16} = \frac{9}{144} \]

Thus:

\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{16}{144} - \frac{9}{144} = \frac{7}{144} \]

Now substituting back into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.0974 \times 10^7 \cdot \frac{7}{144} \]

Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1.0974 \times 10^7 \cdot 7}{144} \]

### Step 3: Calculate the Frequency
We know that the frequency \( \nu \) is related to the wavelength \( \lambda \) by the equation:

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

Where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \).

Since we have \( \frac{1}{\lambda} \), we can express the frequency as:

\[ \nu = R \cdot c \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting the known values:

\[ \nu = 1.0974 \times 10^7 \cdot 3 \times 10^8 \cdot \frac{7}{144} \]

Calculating this gives:

\[ \nu = \frac{3.0002 \times 10^{15}}{144} \approx 2.083 \times 10^{14} \, \text{s}^{-1} \]

### Step 4: Calculate the Energy Difference
The energy difference \( \Delta E \) can be calculated using Planck's equation:

\[ \Delta E = h \cdot \nu \]

Where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J s}) \).

Substituting the values:

\[ \Delta E = 6.626 \times 10^{-34} \cdot 2.083 \times 10^{14} \]

Calculating this gives:

\[ \Delta E \approx 1.38 \times 10^{-19} \, \text{J} \]

### Final Results
- The energy difference \( \Delta E \) is approximately \( 1.38 \times 10^{-19} \, \text{J} \).
- The frequency \( \nu \) is approximately \( 2.083 \times 10^{14} \, \text{s}^{-1} \).