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

Answer

To solve the problem regarding the wavelengths of photons emitted by electron transitions in hydrogen (H) and helium ion (He⁺), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Transition**:
The energy of the photon emitted during an electron transition from a higher energy level (n2) to a lower energy level (n1) can be expressed as:
\[ \Delta E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the emitted photon.

2. **Apply the Formula for Hydrogen**:
For hydrogen (where the atomic number \(Z = 1\)), the energy difference for the transition can be written as:
\[ \Delta E_H = \frac{hc}{\lambda_1} \]

3. **Apply the Formula for Helium Ion (He⁺)**:
For helium ion (where the atomic number \(Z = 2\)), the energy difference for the transition can be expressed as:
\[ \Delta E_{He^+} = \frac{hc}{\lambda_2} = Z^2 \cdot \Delta E_H = 4 \cdot \Delta E_H \]
Therefore, we can write:
\[ \frac{hc}{\lambda_2} = 4 \cdot \frac{hc}{\lambda_1} \]

4. **Cancel out Common Terms**:
Since \(hc\) appears in both equations, we can cancel it out:
\[ \frac{1}{\lambda_2} = 4 \cdot \frac{1}{\lambda_1} \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ \lambda_2 = \frac{\lambda_1}{4} \]

6. **Conclusion**:
This means that the wavelength of the photon emitted by the transition in helium ion is one-fourth the wavelength of the photon emitted in hydrogen:
\[ \lambda_2 = \frac{1}{4} \lambda_1 \]

### Final Answer:
Thus, the relationship between the wavelengths is:
\[ \lambda_2 = \frac{1}{4} \lambda_1 \]