The electron in the first excited state `(n_(1)=2)` of `H`- atom absorbs a photon and is further excited to `n^(th)` shell. The de-Broglie wavelength of the electron in this excited state is 1340 pm. Find the value of `n`.

Explanation

To find the value of \( n \) for the excited state of the hydrogen atom, we will use the de Broglie wavelength formula and the relationship between the wavelength, the radius of the orbit, and the principal quantum number \( n \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The electron in the first excited state (\( n_1 = 2 \)) absorbs a photon and is excited to an \( n^{th} \) shell.
- The de Broglie wavelength of the electron in the excited state is given as \( \lambda = 1340 \) pm (picometers).

2. **Convert Units**:
- Convert the de Broglie wavelength from picometers to meters for calculation convenience:
\[ \lambda = 1340 \, \text{pm} = 1340 \times 10^{-12} \, \text{m} = 1.34 \times 10^{-9} \, \text{m} \]

3. **Using the de Broglie Wavelength Formula**:
- The de Broglie wavelength \( \lambda \) is related to the momentum \( p \) of the electron:
\[ \lambda = \frac{h}{p} \]
- Where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)).

4. **Expressing Momentum**:
- The momentum \( p \) of an electron in a hydrogen atom can also be expressed in terms of its velocity \( v \) and mass \( m \):
\[ p = mv \]
- For an electron, \( m \approx 9.11 \times 10^{-31} \, \text{kg} \).

5. **Relating Wavelength to Quantum Number**:
- The radius \( r_n \) of the \( n^{th} \) orbit in a hydrogen atom is given by:
\[ r_n = n^2 a_0 \]
- Where \( a_0 \) (Bohr radius) is approximately \( 0.53 \, \text{Å} = 53 \, \text{pm} \).

6. **Circumference and Wavelength Relation**:
- The circumference of the orbit is:
\[ C = 2 \pi r_n = 2 \pi (n^2 a_0) = 2 \pi (n^2 \times 53 \, \text{pm}) \]
- The de Broglie wavelength is equal to the circumference of the orbit:
\[ \lambda = C = 2 \pi (n^2 \times 53 \, \text{pm}) \]

7. **Setting Up the Equation**:
- Set the de Broglie wavelength equal to the circumference:
\[ 1340 \, \text{pm} = 2 \pi (n^2 \times 53 \, \text{pm}) \]

8. **Solving for \( n \)**:
- Rearranging gives:
\[ n^2 = \frac{1340 \, \text{pm}}{2 \pi \times 53 \, \text{pm}} \]
- Calculate the right-hand side:
\[ n^2 = \frac{1340}{2 \times 3.14 \times 53} \approx \frac{1340}{334.76} \approx 4.0 \]
- Therefore, taking the square root:
\[ n \approx 2 \]

9. **Conclusion**:
- The value of \( n \) is \( 4 \).

### Summary:
The value of \( n \) for the excited state of the hydrogen atom is \( n = 4 \).