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. Option1 `^^_(n)^(2)~~lambda`
B. Option2 `^^_(n)~~A+(B)/(lambda_(n)^(2))`
C. Option3 `^^_(n)~~A+Blambda_(n)`
D. Option4 `^^_(n)^(2)~~A+B lambda_(n)^(2)`

Answer

To solve the problem, we need to establish relationships between the de Broglie wavelength of the electron in the nth state (\( \lambda_n \)), the de Broglie wavelength in the ground state (\( \lambda_g \)), and the wavelength of the emitted photon (\( \Delta \lambda_n \)) when the electron transitions from the nth state to the ground state.

### Step-by-Step Solution:

1. **Understanding De Broglie Wavelength:**
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its velocity.

2. **Angular Momentum Quantization:**
For the hydrogen atom, the angular momentum of the electron in the nth orbit is quantized and given by:
\[ L = mvr = n\frac{h}{2\pi} \]
where \( r \) is the radius of the nth orbit.

3. **Relating Velocity and Radius:**
From the above equation, we can express \( mv \) as:
\[ mv = \frac{nh}{2\pi r} \]

4. **Substituting into De Broglie Wavelength:**
Substituting \( mv \) into the de Broglie wavelength formula, we get:
\[ \lambda_n = \frac{h}{mv} = \frac{h \cdot 2\pi r}{nh} = \frac{2\pi r}{n} \]

5. **Finding the Radius of nth Orbit:**
The radius of the nth orbit (\( r_n \)) in a hydrogen atom is given by:
\[ r_n = r_0 n^2 \]
where \( r_0 \) is the radius of the ground state.

6. **Substituting for \( r_n \):**
Substituting \( r_n \) into the equation for \( \lambda_n \):
\[ \lambda_n = \frac{2\pi (r_0 n^2)}{n} = 2\pi r_0 n \]

7. **Finding the Wavelength of the Emitted Photon:**
The wavelength of the emitted photon (\( \Delta \lambda_n \)) during the transition from the nth state to the ground state can be expressed using the Rydberg formula:
\[ \frac{1}{\Delta \lambda_n} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \]
where \( R \) is the Rydberg constant.

8. **For Large n:**
As \( n \) becomes very large, we can use the binomial approximation:
\[ \Delta \lambda_n \approx \frac{1}{R} \left( 1 - \frac{1}{n^2} \right) \approx \frac{1}{R} + \frac{1}{R n^2} \]

9. **Relating \( \Delta \lambda_n \) to \( \lambda_n \):**
We know \( \lambda_n = 2\pi r_0 n \). Thus, we can express \( \Delta \lambda_n \) in terms of \( \lambda_n \):
\[ \Delta \lambda_n \approx A + \frac{B}{\lambda_n^2} \]
where \( A \) and \( B \) are constants derived from the Rydberg constant and other factors.

### Final Result:
The relationship we derived is:
\[ \Delta \lambda_n = A + \frac{B}{\lambda_n^2} \]