Consider the spectral line resulting from the transition `n = 2 rarr n = 1 ` in the atoms and ions given . The shortest wavelength is produced by

Options

A. hydrogen atom
B. deuterium atom
C. singly ionization helium
D. doubly ionized lithium

Explanation

To solve the problem of determining which atom or ion produces the shortest wavelength for the transition from \( n = 2 \) to \( n = 1 \), we can follow these steps:

### Step 1: Understand the formula for wavelength
The wavelength of the spectral line resulting from an electronic transition can be calculated using the formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.

### Step 2: Substitute the values for the transition
For the transition from \( n = 2 \) to \( n = 1 \):
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the formula gives:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = RZ^2 \left( 1 - \frac{1}{4} \right) = RZ^2 \left( \frac{3}{4} \right) \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{1}{\lambda} = \frac{3RZ^2}{4} \]
From this equation, we can see that \( \lambda \) is inversely proportional to \( Z^2 \):
\[ \lambda \propto \frac{1}{Z^2} \]
This means that as the atomic number \( Z \) increases, the wavelength \( \lambda \) decreases.

### Step 4: Identify the atomic numbers of the options
Now, we need to identify the atomic numbers of the given options:
1. Hydrogen (H) - \( Z = 1 \)
2. Deuterium (D) - \( Z = 1 \) (isotope of Hydrogen)
3. Singly ionized Helium (He\(^+\)) - \( Z = 2 \)
4. Doubly ionized Lithium (Li\(^{2+}\)) - \( Z = 3 \)

### Step 5: Determine which has the greatest atomic number
The atomic numbers are:
- Hydrogen: \( Z = 1 \)
- Deuterium: \( Z = 1 \)
- Singly ionized Helium: \( Z = 2 \)
- Doubly ionized Lithium: \( Z = 3 \)

The greatest atomic number is \( Z = 3 \) for doubly ionized Lithium.

### Step 6: Conclusion
Since the wavelength is inversely proportional to \( Z^2 \), the atom or ion with the highest atomic number will produce the shortest wavelength. Therefore, the shortest wavelength is produced by:

**Doubly ionized Lithium (Li\(^{2+}\))**.

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