At what atomic number would a transition from `n=2"to"n=1` energy level result in emission of photon of `lambda= 3xx10^(-8)m`?

Answer

To find the atomic number (Z) at which a transition from the energy level n=2 to n=1 results in the emission of a photon with a wavelength of \( \lambda = 3 \times 10^{-8} \) m, we can use the Rydberg formula for hydrogen-like atoms:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R \) is the Rydberg constant (\( R \approx 1.0967 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.

### Step 1: Identify the values for \( n_1 \) and \( n_2 \)
For this transition:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

### Step 2: Substitute the known values into the Rydberg formula
We can rewrite the formula with the known values:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

### Step 3: Calculate \( \frac{1}{1^2} - \frac{1}{2^2} \)
Calculating the right side:
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 4: Substitute \( \lambda \) and \( R \) into the equation
Now substituting \( \lambda = 3 \times 10^{-8} \) m and \( R = 1.0967 \times 10^7 \, \text{m}^{-1} \):
\[ \frac{1}{3 \times 10^{-8}} = 1.0967 \times 10^7 \cdot Z^2 \cdot \frac{3}{4} \]

### Step 5: Calculate \( \frac{1}{3 \times 10^{-8}} \)
Calculating the left side:
\[ \frac{1}{3 \times 10^{-8}} \approx 3.33 \times 10^7 \, \text{m}^{-1} \]

### Step 6: Rearranging the equation to solve for \( Z^2 \)
Now we can rearrange the equation:
\[ 3.33 \times 10^7 = 1.0967 \times 10^7 \cdot Z^2 \cdot \frac{3}{4} \]

### Step 7: Isolate \( Z^2 \)
\[ Z^2 = \frac{3.33 \times 10^7}{1.0967 \times 10^7 \cdot \frac{3}{4}} \]

### Step 8: Calculate the value of \( Z^2 \)
Calculating the right side:
\[ Z^2 = \frac{3.33}{1.0967 \cdot \frac{3}{4}} \approx \frac{3.33}{0.822525} \approx 4.05 \]

### Step 9: Take the square root to find \( Z \)
\[ Z = \sqrt{4.05} \approx 2 \]

### Conclusion
The atomic number \( Z \) at which the transition from \( n=2 \) to \( n=1 \) results in the emission of a photon of wavelength \( 3 \times 10^{-8} \) m is approximately **2**.

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