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 solve the problem of determining the atomic number (Z) at which a transition from the n=2 to n=1 energy level results in the emission of a photon with a wavelength of λ = 3 x 10^(-8) m, we can use the Rydberg formula for hydrogen-like atoms. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wavelength of emitted or absorbed light during electronic transitions in hydrogen-like atoms is given by:

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

Where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R \) is the Rydberg constant (\( R \approx 1.09 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states (with \( n_2 > n_1 \)).

### Step 2: Substitute Known Values
In this case:
- \( n_1 = 1 \)
- \( n_2 = 2 \)
- \( \lambda = 3 \times 10^{-8} \, \text{m} \)

Substituting these values into the formula gives:

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

### Step 3: Calculate the Right Side
Calculate \( \frac{1}{1^2} - \frac{1}{2^2} \):

\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 4: Substitute into the Equation
Now substituting this back into the equation:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{3}{4} \right) \]

### Step 5: Calculate \( \frac{1}{\lambda} \)
Calculate \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1}{3 \times 10^{-8}} = \frac{1}{3} \times 10^{8} \, \text{m}^{-1} \approx 3.33 \times 10^{7} \, \text{m}^{-1} \]

### Step 6: Set Up the Equation
Now we have:

\[ 3.33 \times 10^{7} = 1.09 \times 10^7 Z^2 \left( \frac{3}{4} \right) \]

### Step 7: Solve for \( Z^2 \)
Rearranging gives:

\[ Z^2 = \frac{3.33 \times 10^{7}}{1.09 \times 10^7 \times \frac{3}{4}} \]

Calculating the denominator:

\[ 1.09 \times 10^7 \times \frac{3}{4} = 8.175 \times 10^6 \]

Now substituting back:

\[ Z^2 = \frac{3.33 \times 10^{7}}{8.175 \times 10^6} \approx 4.07 \]

### Step 8: Calculate \( Z \)
Taking the square root gives:

\[ Z \approx \sqrt{4.07} \approx 2.02 \]

Since atomic numbers are whole numbers, we round this to:

\[ Z = 2 \]

### Final Answer
The atomic number at which a transition from \( n=2 \) to \( n=1 \) results in the emission of a photon with a wavelength of \( \lambda = 3 \times 10^{-8} \, \text{m} \) is:

\[ \boxed{2} \]