Maximum wavelength of lyman series photon for H is then minimum wavelength of Ballmer series photon for `He` atom.

Options

A. `lambda/4`
B. ``3lambda/4`
C. `lambda/4`
D. `2lamda/3`

Answer

To solve the problem, we need to find the maximum wavelength of the Lyman series photon for a hydrogen atom and then relate it to the minimum wavelength of the Balmer series photon for a helium atom.

### Step 1: Understanding the Lyman Series for Hydrogen
The Lyman series corresponds to electronic transitions in a hydrogen atom where the electron falls to the n=1 energy level. The maximum wavelength in this series occurs when the electron transitions from n=2 to n=1.

### Step 2: Using the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during an electron transition is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]
Where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (for Lyman series, \( n_1 = 1 \)),
- \( n_2 \) is the higher energy level (for maximum wavelength, \( n_2 = 2 \)).

### Step 3: Calculate the Maximum Wavelength for Hydrogen
Substituting the values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) \]
\[ \frac{1}{\lambda} = R \left(1 - \frac{1}{4}\right) = R \left(\frac{3}{4}\right) \]
\[ \lambda = \frac{4}{3R} \]

### Step 4: Understanding the Balmer Series for Helium
The Balmer series corresponds to transitions where the electron falls to the n=2 energy level. The minimum wavelength occurs when the electron transitions from n=∞ to n=2.

### Step 5: Calculate the Minimum Wavelength for Helium
For a helium atom (specifically He+), the atomic number \( Z = 2 \). Using the Rydberg formula again:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]
For the minimum wavelength in the Balmer series:
\[ \frac{1}{\lambda} = R \cdot 2^2 \left(\frac{1}{2^2} - 0\right) = R \cdot 4 \cdot \frac{1}{4} = R \]
Thus, we have:
\[ \lambda = \frac{1}{R} \]

### Step 6: Relating the Two Wavelengths
Now, we need to relate the maximum wavelength of the Lyman series for hydrogen to the minimum wavelength of the Balmer series for helium:
\[ \lambda_{\text{He}} = \frac{1}{R} \]
\[ \lambda_{\text{H}} = \frac{4}{3R} \]
To find the relationship:
\[ \lambda_{\text{He}} = \frac{3}{4} \lambda_{\text{H}} \]
Thus, the minimum wavelength of the Balmer series for helium is \( \frac{3}{4} \) times the maximum wavelength of the Lyman series for hydrogen.

### Final Answer
The maximum wavelength of the Lyman series photon for hydrogen is then \( \frac{3}{4} \) times the minimum wavelength of the Balmer series photon for helium.