Consider a spectral line resulting from the transition n = 5 to n = 1, in the atoms and ions given below. The shortest wavelength is produced by

Options

A. helium atom
B. deuterium atom
C. singly ionized helium
D. ten times ionized sodium atom

Answer

To determine which atom or ion produces the shortest wavelength for the transition from n = 5 to n = 1, we can follow these steps:

### Step 1: Understand the formula for wavelength
The wavelength (\(\lambda\)) of the spectral line can be calculated using the formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels, respectively.

### Step 2: Identify the transition
In this case, the transition is from \(n = 5\) to \(n = 1\). Thus, we have:
- \(n_1 = 1\)
- \(n_2 = 5\)

### Step 3: Calculate the wavelength dependency on atomic number
From the formula, we can see that the wavelength is inversely proportional to \(Z^2\). This means that as the atomic number \(Z\) increases, the wavelength \(\lambda\) decreases. Therefore, the atom or ion with the largest atomic number will produce the shortest wavelength.

### Step 4: Analyze the given options
The options provided are:
1. Helium atom (Z = 2)
2. Deuterium atom (Z = 1)
3. Singly ionized helium (He\(^+\), Z = 2)
4. 10 times ionized sodium (Na\(^+\) with 10 electrons removed, Z = 11)

### Step 5: Determine which has the largest atomic number
Among the options:
- Helium (Z = 2)
- Deuterium (Z = 1)
- Singly ionized helium (Z = 2)
- 10 times ionized sodium (Z = 11)

The 10 times ionized sodium has the largest atomic number \(Z = 11\).

### Step 6: Conclusion
Since the 10 times ionized sodium has the largest atomic number, it will produce the shortest wavelength for the transition from \(n = 5\) to \(n = 1\).

**Final Answer:** The shortest wavelength is produced by the 10 times ionized sodium atom.

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