An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let `lambda_(n),lambda_(g)` be the de Broglie wavelength of the electron in the `n^(th)` state and the ground state respectively. Let `^^^_(n)` be the wavelength of the emitted photon in the transition from the `n^(th)` state to the ground state. For large n, (A, B are constants)

Options

A. `^^_(n)^(2)~~lambda`
B. `^^_(n)~~A+(B)/(lambda_(n)^(2))`
C. `^^_(n)~~A+Blambda_(n)`
D. `^^_(n)^(2)~~A+B lambda_(n)^(2)`

Answer

To solve the problem, we need to analyze the relationship between the de Broglie wavelength of an electron in the nth excited state of a hydrogen atom and the wavelength of the photon emitted when the electron transitions to the ground state.

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda_n \)) of an electron in the nth state can be expressed as:
\[ \lambda_n = \frac{2\pi r}{n} \]
where \( r \) is the radius of the electron's orbit in the nth state.

2. **Radius of the Electron's Orbit**:
For a hydrogen atom, the radius of the nth orbit is given by:
\[ r_n = n^2 a_0 \]
where \( a_0 \) is the Bohr radius. Therefore, substituting \( r_n \) into the de Broglie wavelength:
\[ \lambda_n = \frac{2\pi (n^2 a_0)}{n} = 2\pi n a_0 \]

3. **Photon Wavelength during Transition**:
When the electron transitions from the nth state to the ground state (n=1), the wavelength of the emitted photon (\( \lambda_{photon} \)) can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda_{photon}} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \]
where \( R \) is the Rydberg constant.

4. **Simplifying the Photon Wavelength**:
Rearranging the above equation gives:
\[ \lambda_{photon} = \frac{1}{R \left( 1 - \frac{1}{n^2} \right)} = \frac{n^2}{R(n^2 - 1)} \]

5. **Finding the Relationship**:
Now we need to relate \( \lambda_{photon} \) to \( \lambda_n \). We can express \( \lambda_{photon} \) in terms of \( \lambda_n \):
\[ \frac{1}{\lambda_{photon}} = R \left( 1 - \frac{1}{n^2} \right) = R \left( \frac{n^2 - 1}{n^2} \right) \]
This implies:
\[ \lambda_{photon} = \frac{n^2}{R(n^2 - 1)} \]

6. **Final Relation**:
Now we can combine the two wavelengths:
\[ \frac{1}{\lambda_{photon}} = \frac{1}{\lambda_n} + \text{constant terms} \]
This can be expressed in the form:
\[ \frac{1}{\lambda_{photon}} = A + \frac{B}{\lambda_n^2} \]
where \( A \) and \( B \) are constants derived from the previous equations.

### Conclusion:
The final relation shows how the wavelength of the emitted photon relates to the de Broglie wavelength of the electron in the nth state.