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 involved in the transition of a hydrogen atom that emits ultraviolet radiation of wavelength 102.5 nm, we can follow these steps:

### Step 1: Identify the Series
The emission of ultraviolet radiation indicates that the transition is likely part of the Lyman series, which involves transitions to the n=1 energy level. In the Lyman series, the final state (n1) is 1.

### Step 2: Write the Formula
We will use the Rydberg formula for hydrogen to find the initial quantum number (n2) involved in the transition:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \(\lambda\) is the wavelength of the emitted light (in meters),
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(n_1\) is the lower energy level (which is 1 for the Lyman series),
- \(n_2\) is the higher energy level we need to find.

### Step 3: Convert Wavelength to Meters
Convert the wavelength from nanometers to meters:
\[ \lambda = 102.5 \, \text{nm} = 102.5 \times 10^{-9} \, \text{m} \]

### Step 4: Substitute Known Values into the Formula
Substituting \(\lambda\) and \(n_1\) into the Rydberg formula:
\[ \frac{1}{102.5 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right) \]

### Step 5: Solve for \(n_2\)
Rearranging the equation gives:
\[ \frac{1}{102.5 \times 10^{-9}} = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right) \]
Calculating the left side:
\[ \frac{1}{102.5 \times 10^{-9}} \approx 9.76 \times 10^6 \]
Now we can set up the equation:
\[ 9.76 \times 10^6 = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right) \]
Dividing both sides by \(1.097 \times 10^7\):
\[ \frac{9.76 \times 10^6}{1.097 \times 10^7} = 1 - \frac{1}{n_2^2} \]
Calculating the left side:
\[ \approx 0.889 \]
Thus, we have:
\[ 0.889 = 1 - \frac{1}{n_2^2} \]
Rearranging gives:
\[ \frac{1}{n_2^2} = 1 - 0.889 = 0.111 \]
Taking the reciprocal:
\[ n_2^2 = \frac{1}{0.111} \approx 9 \]
Taking the square root:
\[ n_2 \approx 3 \]

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