The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of wavelength `lambda_(0) = (16)/(15R)`, where R is Rydbergs's constant. In place of emitting one photon, the electron could come back to ground state by

Options

A. a.Emitting 3 photons of wavelengths `lambda_1,lambda_2` and `lambda_3` such that `1/lambda_1+1/lambda_2+1/lambda_3="15R"/16`
B. b.Emitting 2 photons of wavelengths `lambda_1` and `lambda_2` such that `1/lambda_1+1/lambda_2="15R"/16`
C. c.Emitting 2 photons of wavelengths `lambda_1` and `lambda_2` such that `lambda_1+lambda_2="15R"/16`
D. d.Emitting 3 photons of wavelengths `lambda_1,lambda_2` and `lambda_3` such that `lambda_1+lambda_2+lambda_3="16"/"15R"`

Answer

To solve the problem of an electron in a hydrogen atom transitioning from an excited state to the ground state by emitting a photon of wavelength \(\lambda_0 = \frac{16}{15R}\), where \(R\) is Rydberg's constant, we can analyze the situation step by step.

### Step 1: Understanding the Transition
The electron can transition from a higher energy level (excited state) to a lower energy level (ground state) by emitting a photon. The wavelength of the emitted photon is related to the energy difference between the two states.

### Step 2: Using Rydberg's Formula
Rydberg's formula for the wavelength of emitted light when an electron transitions between two energy levels is given by:

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

where \(n_1\) is the principal quantum number of the lower energy level (ground state) and \(n_2\) is the principal quantum number of the higher energy level (excited state).

### Step 3: Identifying the Quantum States
In our case, we can assume \(n_1 = 1\) (ground state) and \(n_2\) is the excited state. We can rearrange Rydberg's formula to find \(n_2\):

\[ \frac{1}{\lambda_0} = R \left( 1 - \frac{1}{n_2^2} \right) \]

Given \(\lambda_0 = \frac{16}{15R}\), we can substitute this into the equation:

\[ \frac{15R}{16} = R \left( 1 - \frac{1}{n_2^2} \right) \]

### Step 4: Simplifying the Equation
Cancelling \(R\) from both sides (assuming \(R \neq 0\)) gives:

\[ \frac{15}{16} = 1 - \frac{1}{n_2^2} \]

Rearranging this yields:

\[ \frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16} \]

### Step 5: Solving for \(n_2\)
Taking the reciprocal gives:

\[ n_2^2 = 16 \implies n_2 = 4 \]

Thus, the electron is transitioning from \(n_2 = 4\) to \(n_1 = 1\).

### Step 6: Considering Alternative Transition Paths
The electron can return to the ground state in multiple ways:
1. **Directly from \(n=4\) to \(n=1\)** (emitting one photon).
2. **In two steps**:
- From \(n=4\) to \(n=2\) and then from \(n=2\) to \(n=1\) (emitting two photons).
- From \(n=4\) to \(n=3\) and then from \(n=3\) to \(n=1\) (emitting two photons).
3. **In three steps**:
- From \(n=4\) to \(n=3\), then \(n=3\) to \(n=2\), and finally \(n=2\) to \(n=1\) (emitting three photons).

### Step 7: Formulating the Relationships
For the two-photon case, we can express the relationship as:

\[ \frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{15R}{16} \]

For the three-photon case, the relationship becomes:

\[ \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{15R}{16} \]

### Conclusion
Thus, the electron can return to the ground state by either emitting two photons or three photons, leading us to the conclusion that the correct answers are:

- Emitting two photons such that \(\frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{15R}{16}\).
- Emitting three photons such that \(\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{15R}{16}\).