What hydrogen-like ion has the wavelength difference between the first lines of the Balmer Lyman series equal to `59.3nm`?

Answer

To solve the problem of finding the hydrogen-like ion that has a wavelength difference between the first lines of the Balmer and Lyman series equal to 59.3 nm, we can follow these steps:

### Step 1: Write down the given data
We know that the wavelength difference between the first lines of the Balmer and Lyman series is given as:
\[ \lambda_B - \lambda_L = 59.3 \, \text{nm} = 59.3 \times 10^{-9} \, \text{m} = 0.593 \times 10^{-6} \, \text{m} \]

### Step 2: Write the wavelength equations
For the first line of the Balmer series (n=3 to n=2):
\[ \frac{1}{\lambda_B} = Z^2 R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the terms:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda_B} = Z^2 R_H \cdot \frac{5}{36} \]
So,
\[ \lambda_B = \frac{36}{5 Z^2 R_H} \]

For the first line of the Lyman series (n=2 to n=1):
\[ \frac{1}{\lambda_L} = Z^2 R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Calculating the terms:
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ \frac{1}{\lambda_L} = Z^2 R_H \cdot \frac{3}{4} \]
So,
\[ \lambda_L = \frac{4}{3 Z^2 R_H} \]

### Step 3: Set up the equation for the wavelength difference
Now, we can express the difference between the two wavelengths:
\[ \lambda_B - \lambda_L = \frac{36}{5 Z^2 R_H} - \frac{4}{3 Z^2 R_H} = 0.593 \times 10^{-6} \, \text{m} \]

### Step 4: Combine the fractions
To combine the fractions, we need a common denominator:
\[ \lambda_B - \lambda_L = \left( \frac{36 \cdot 3 - 4 \cdot 5}{15 Z^2 R_H} \right) = \frac{108 - 20}{15 Z^2 R_H} = \frac{88}{15 Z^2 R_H} \]
Setting this equal to the wavelength difference:
\[ \frac{88}{15 Z^2 R_H} = 0.593 \times 10^{-6} \, \text{m} \]

### Step 5: Solve for \( Z^2 \)
Rearranging gives:
\[ Z^2 = \frac{88}{15 R_H \cdot 0.593 \times 10^{-6}} \]
Using \( R_H = 1.09678 \times 10^5 \, \text{m}^{-1} \):
\[ Z^2 = \frac{88}{15 \cdot 1.09678 \times 10^5 \cdot 0.593 \times 10^{-6}} \]
Calculating the right-hand side:
\[ Z^2 = \frac{88}{15 \cdot 1.09678 \cdot 0.593 \cdot 10^{-1}} \approx 9 \]

### Step 6: Find \( Z \)
Taking the square root gives:
\[ Z = \sqrt{9} = 3 \]

### Step 7: Identify the ion
The atomic number \( Z = 3 \) corresponds to lithium (Li). Since we are looking for a hydrogen-like ion, the ion is \( \text{Li}^{2+} \).

### Final Answer
The hydrogen-like ion with the specified wavelength difference is \( \text{Li}^{2+} \).

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