The wavelengths of photons emitted by electron transition between two similar leveis in H and `He^(+)` are `lambda_(1)` and `lambda_(2)` respectively. Then :-

Options

A. `lambda_(2)=lambda_(1)`
B. `lambda_(2)=2lambda_(1)`
C. `lambda_(2)=lambda_(1)//2`
D. `lambda_(2)=lambda_(1)//4`

Explanation

To solve the problem regarding the wavelengths of photons emitted by electron transitions in hydrogen (H) and helium ion (He⁺), we need to analyze the energy levels and the relationship between energy and wavelength.

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{E_1 \cdot Z^2}{n^2} \]
where \(E_1\) is the energy of the ground state, \(Z\) is the atomic number, and \(n\) is the principal quantum number.

2. **Energy of Photons**:
The energy of a photon emitted during a transition is also related to its wavelength (\(\lambda\)) by the equation:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light.

3. **Relating the Two Atoms**:
For hydrogen (H), \(Z = 1\) and for helium ion (He⁺), \(Z = 2\). Therefore, the energy levels for both can be expressed as:
- For hydrogen:
\[ E_n^{H} = -\frac{E_1 \cdot 1^2}{n^2} = -\frac{E_1}{n^2} \]
- For helium ion:
\[ E_n^{He^+} = -\frac{E_1 \cdot 2^2}{n^2} = -\frac{4E_1}{n^2} \]

4. **Setting Up the Relationship**:
The energy of the emitted photon during a transition can be expressed as:
\[ E = E_{\text{initial}} - E_{\text{final}} \]
Therefore, for transitions in hydrogen and helium ion, we can write:
\[ E_{\text{photon}}^{H} = E_n^{H} \quad \text{and} \quad E_{\text{photon}}^{He^+} = E_n^{He^+} \]

5. **Finding the Wavelengths**:
From the energy-wavelength relationship, we can derive:
\[ \lambda_1 \propto \frac{1}{E_n^{H}} \quad \text{and} \quad \lambda_2 \propto \frac{1}{E_n^{He^+}} \]

6. **Calculating the Ratio of Wavelengths**:
The ratio of the wavelengths can be expressed as:
\[ \frac{\lambda_2}{\lambda_1} = \frac{E_n^{H}}{E_n^{He^+}} = \frac{1^2}{2^2} = \frac{1}{4} \]
Thus, we can conclude:
\[ \lambda_2 = 4 \lambda_1 \]

### Final Result:
The relationship between the wavelengths of photons emitted by electron transitions in hydrogen and helium ion is:
\[ \lambda_2 = 4 \lambda_1 \]