What electron transition in the `He^(+)` spectrum would have the same wavelength as the first Lyman transition of hydrogen.

Answer

To solve the problem of determining which electron transition in the \( \text{He}^+ \) spectrum has the same wavelength as the first Lyman transition of hydrogen, we will follow these steps:

### Step 1: Understand the Lyman Transition of Hydrogen
The first Lyman transition in hydrogen occurs when an electron transitions from the second energy level (n=2) to the first energy level (n=1). The energy of this transition can be calculated using the formula for the energy levels of a hydrogen-like atom:

\[ E_n = -\frac{R_H}{n^2} \]

Where \( R_H \) (Rydberg constant for hydrogen) is approximately 13.6 eV.

### Step 2: Calculate the Energy for the First Lyman Transition in Hydrogen
For the first Lyman transition (n=2 to n=1):

\[ E_H = E_1 - E_2 = -\frac{R_H}{1^2} - \left(-\frac{R_H}{2^2}\right) = -13.6 eV - \left(-\frac{13.6 eV}{4}\right) \]

Calculating this gives:

\[ E_H = -13.6 eV + 3.4 eV = -10.2 eV \]

### Step 3: Determine the Energy Levels for \( \text{He}^+ \)
For \( \text{He}^+ \), which is a hydrogen-like ion with \( Z = 2 \), the energy levels are given by:

\[ E_n = -\frac{R_{He}}{n^2} \]

Where \( R_{He} = Z^2 \cdot R_H = 4 \cdot 13.6 eV = 54.4 eV \).

### Step 4: Set Up the Equation for the Transition in \( \text{He}^+ \)
We need to find a transition in \( \text{He}^+ \) that has the same energy as the first Lyman transition in hydrogen.

Let’s denote the transition in \( \text{He}^+ \) from \( n_i \) to \( n_f \):

\[ E_{He} = E_{n_f} - E_{n_i} = -\frac{R_{He}}{n_f^2} + \frac{R_{He}}{n_i^2} \]

### Step 5: Set the Energies Equal
To find the transition that matches the energy of the first Lyman transition of hydrogen:

\[ -\frac{R_{He}}{n_f^2} + \frac{R_{He}}{n_i^2} = -10.2 eV \]

Substituting \( R_{He} = 54.4 eV \):

\[ -\frac{54.4}{n_f^2} + \frac{54.4}{n_i^2} = -10.2 \]

### Step 6: Simplify and Solve
Rearranging gives:

\[ \frac{54.4}{n_i^2} - \frac{54.4}{n_f^2} = 10.2 \]

Dividing through by 54.4:

\[ \frac{1}{n_i^2} - \frac{1}{n_f^2} = \frac{10.2}{54.4} \]

Calculating \( \frac{10.2}{54.4} \approx 0.1875 \).

### Step 7: Find Suitable \( n_i \) and \( n_f \)
We can test various values for \( n_i \) and \( n_f \):

Assuming \( n_f = 2 \) (as it is the lower level):

\[ \frac{1}{n_i^2} - \frac{1}{4} = 0.1875 \]

Solving for \( n_i \):

\[ \frac{1}{n_i^2} = 0.1875 + 0.25 = 0.4375 \]

Thus,

\[ n_i^2 = \frac{1}{0.4375} \approx 2.2857 \implies n_i \approx 4.8 \]

Since \( n_i \) must be an integer, we take \( n_i = 4 \).

### Conclusion
The electron transition in the \( \text{He}^+ \) spectrum that has the same wavelength as the first Lyman transition of hydrogen is from \( n = 4 \) to \( n = 2 \).