An electron in the first excited state of H atom absorbed a photon and further excited. The de broglie wavelength of the electron in this state is found to be `13.4` Å. Find the wavelength of the photon absorbed by the electron in angstroms. Also find the longest and the shortest wavelength emitted when this electron de-excited back to the ground state.

Explanation

To solve the problem step by step, we will first determine the wavelength of the photon absorbed by the electron when it transitions from the first excited state (n=2) to a higher energy level (n=4). Then, we will find the longest and shortest wavelengths emitted when the electron de-excites back to the ground state (n=1).

### Step 1: Determine the value of n for the excited state
Given the de Broglie wavelength (λ) of the electron in the excited state is 13.4 Å, we can use the formula derived from Bohr's model:

\[ 2\pi r = n \lambda \]

Where \( r \) is the radius of the electron's orbit given by:

\[ r = \frac{n^2}{Z} \cdot 0.529 \, \text{Å} \]

For hydrogen, \( Z = 1 \). Therefore, we can rewrite the equation as:

\[ 2\pi \left( \frac{n^2}{1} \cdot 0.529 \right) = n \cdot 13.4 \]

This simplifies to:

\[ 2\pi \cdot 0.529 n^2 = n \cdot 13.4 \]

Dividing both sides by \( n \) (assuming \( n \neq 0 \)):

\[ 2\pi \cdot 0.529 n = 13.4 \]

Now, solving for \( n \):

\[ n = \frac{13.4}{2\pi \cdot 0.529} \]

Calculating this gives:

\[ n \approx 4 \]

### Step 2: Find the wavelength of the absorbed photon
To find the wavelength of the photon absorbed when the electron transitions from n=2 to n=4, we use the Rydberg formula:

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

Where:
- \( R_H \) (Rydberg constant) = \( \frac{1}{912} \, \text{Å} \)
- \( n_L = 2 \) (lower energy level)
- \( n_H = 4 \) (higher energy level)

Substituting the values:

\[ \frac{1}{\lambda} = \frac{1}{912} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda} = \frac{1}{912} \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{1}{912} \left( \frac{4 - 1}{16} \right) = \frac{3}{16 \cdot 912} \]

Calculating \( \lambda \):

\[ \lambda = \frac{16 \cdot 912}{3} \approx 4.86 \times 10^3 \, \text{Å} \]

### Step 3: Find the longest and shortest wavelengths emitted
When the electron de-excites from n=4 to n=1, it can do so in multiple steps. The longest wavelength corresponds to the transition from n=4 to n=3, and the shortest wavelength corresponds to the transition from n=4 to n=1.

**Longest Wavelength (n=4 to n=3):**

Using the Rydberg formula again:

\[ \frac{1}{\lambda_{max}} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

Calculating:

\[ \frac{1}{\lambda_{max}} = \frac{1}{912} \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{1}{912} \left( \frac{16 - 9}{144} \right) = \frac{7}{144 \cdot 912} \]

Calculating \( \lambda_{max} \):

\[ \lambda_{max} = \frac{144 \cdot 912}{7} \approx 1.87 \times 10^4 \, \text{Å} \]

**Shortest Wavelength (n=4 to n=1):**

Using the Rydberg formula:

\[ \frac{1}{\lambda_{min}} = R_H \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]

Calculating:

\[ \frac{1}{\lambda_{min}} = \frac{1}{912} \left( 1 - \frac{1}{16} \right) = \frac{1}{912} \left( \frac{15}{16} \right) = \frac{15}{16 \cdot 912} \]

Calculating \( \lambda_{min} \):

\[ \lambda_{min} = \frac{16 \cdot 912}{15} \approx 855 \, \text{Å} \]

### Final Answers:
- Wavelength of the absorbed photon: \( \approx 4860 \, \text{Å} \)
- Longest wavelength emitted: \( \approx 1.87 \times 10^4 \, \text{Å} \)
- Shortest wavelength emitted: \( \approx 855 \, \text{Å} \)