A hydrogen atom emits ultraviolet of wavelength `102.5 nm` what are the quantum number of the state involved in the transition?

Explanation

To determine the quantum numbers of the states involved in the transition of a hydrogen atom that emits ultraviolet radiation of wavelength \(102.5 \, \text{nm}\), we can follow these steps:

### Step 1: Convert Wavelength to Meters
First, we convert the wavelength from nanometers to meters:
\[ \lambda = 102.5 \, \text{nm} = 102.5 \times 10^{-9} \, \text{m} \]

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light in hydrogen is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)),
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels, respectively, with \(n_2 > n_1\).

### Step 3: Calculate \( \frac{1}{\lambda} \)
Substituting the value of \(\lambda\):
\[ \frac{1}{\lambda} = \frac{1}{102.5 \times 10^{-9}} \approx 9.76 \times 10^6 \, \text{m}^{-1} \]

### Step 4: Substitute into the Rydberg Formula
Now, substituting into the Rydberg formula:
\[ 9.76 \times 10^6 = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 5: Simplify the Equation
Dividing both sides by \(1.097 \times 10^7\):
\[ \frac{9.76 \times 10^6}{1.097 \times 10^7} \approx 0.889 = \frac{1}{n_1^2} - \frac{1}{n_2^2} \]

### Step 6: Identify the Series
Since the emitted radiation is ultraviolet, it corresponds to the Lyman series, which involves transitions to \(n_1 = 1\). Thus, we set \(n_1 = 1\).

### Step 7: Solve for \(n_2\)
Substituting \(n_1 = 1\) into the equation:
\[ 0.889 = 1 - \frac{1}{n_2^2} \]
\[ \frac{1}{n_2^2} = 1 - 0.889 = 0.111 \]
\[ n_2^2 = \frac{1}{0.111} \approx 9 \]
\[ n_2 = 3 \]

### Conclusion
The quantum numbers involved in the transition are:
- \(n_1 = 1\) (lower energy level)
- \(n_2 = 3\) (upper energy level)