The longest wavelength that the singly ionized helium atom in its ground state will obsorb is

Options

A. `912 Å`
B. `304 Å`
C. `606 Å`
D. `1216 Å`

Explanation

To find the longest wavelength that the singly ionized helium atom (He+) in its ground state will absorb, we can follow these steps:

### Step 1: Identify the transition levels
For the longest wavelength absorbed, the transition occurs from the second energy level (n_i = 2) to the first energy level (n_f = 1).

### Step 2: Use the formula for wavelength
The formula relating the wavelength (λ) to the energy levels is given by:

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \( R \) is the Rydberg constant, approximately \( 912 \, \text{Å}^{-1} \) (in terms of Angstroms).
- \( Z \) is the atomic number (for singly ionized helium, \( Z = 2 \)).
- \( n_f \) is the final energy level (1 in this case).
- \( n_i \) is the initial energy level (2 in this case).

### Step 3: Substitute the values into the formula
Substituting the known values into the equation:

\[ \frac{1}{\lambda} = 912 \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

### Step 4: Calculate the terms
Calculating the right-hand side:

1. Calculate \( Z^2 \):
\[ Z^2 = 2^2 = 4 \]

2. Calculate \( \frac{1}{n_f^2} - \frac{1}{n_i^2} \):
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

3. Substitute these values back into the equation:
\[ \frac{1}{\lambda} = 912 \cdot 4 \cdot \frac{3}{4} \]
Simplifying:
\[ \frac{1}{\lambda} = 912 \cdot 3 = 2736 \]

### Step 5: Calculate λ
Now, take the reciprocal to find λ:
\[ \lambda = \frac{1}{2736} \text{ Å}^{-1} \]

To convert this to Angstroms:
\[ \lambda = \frac{1}{2736} \approx 0.000365 \text{ m} = 365 \text{ nm} = 3650 \text{ Å} \]

### Step 6: Final answer
Thus, the longest wavelength that the singly ionized helium atom in its ground state will absorb is approximately:

\[ \lambda \approx 304 \, \text{Å} \]