A hydrogen atom initially in the ground level absorbs a photon and is excited to `n=4` level then the wavelength of photon is

Options

A. `790 Å`
B. `870 Å`
C. `970 Å`
D. `1070 Å`

Explanation

To find the wavelength of the photon absorbed by a hydrogen atom to excite it from the ground state (n=1) to the n=4 level, we can use the Rydberg formula for hydrogen:

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

Where:
- \( \lambda \) is the wavelength of the emitted or absorbed photon,
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (initial state),
- \( n_2 \) is the higher energy level (final state).

### Step-by-step Solution:

1. **Identify the initial and final energy levels**:
- The hydrogen atom starts at the ground state, so \( n_1 = 1 \).
- The atom is excited to the \( n=4 \) level, so \( n_2 = 4 \).

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]

3. **Calculate the terms**:
- \( \frac{1}{1^2} = 1 \)
- \( \frac{1}{4^2} = \frac{1}{16} \)

4. **Combine the fractions**:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{16}{16} - \frac{1}{16} \right) = R \left( \frac{15}{16} \right) \]

5. **Substitute the value of R**:
\[ \frac{1}{\lambda} = R \cdot \frac{15}{16} = 1.097 \times 10^7 \cdot \frac{15}{16} \]

6. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot \frac{15}{16} = 1.097 \times 10^7 \cdot 0.9375 \approx 1.028 \times 10^7 \, \text{m}^{-1} \]

7. **Find \( \lambda \)**:
\[ \lambda = \frac{1}{1.028 \times 10^7} \approx 9.73 \times 10^{-8} \, \text{m} \]

8. **Convert to Angstroms**:
- Since \( 1 \, \text{m} = 10^{10} \, \text{Å} \):
\[ \lambda \approx 9.73 \times 10^{-8} \, \text{m} \times 10^{10} \, \text{Å/m} \approx 973 \, \text{Å} \]

### Final Answer:
The wavelength of the photon absorbed by the hydrogen atom is approximately \( 970 \, \text{Å} \).