The wavelength of high energy transition of H atom is `91.2 nm` Calculate the corresponding wavelength of He atom

Explanation

To calculate the corresponding wavelength of the He atom given the wavelength of a high energy transition of the H atom, we can follow these steps:

### Step 1: Understand the formula for wavelength in hydrogen and helium ions
For hydrogen (H), the formula for the wavelength of a transition is given by:
\[ \frac{1}{\lambda} = R_H z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \(\lambda\) is the wavelength,
- \(R_H\) is the Rydberg constant for hydrogen,
- \(z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels.

### Step 2: Write the equation for the He+ ion
For the He+ ion, the equation is similar:
\[ \frac{1}{\lambda_{He^+}} = R_H z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
In this case, \(z\) for He+ is 2.

### Step 3: Relate the wavelengths of H and He+
From the equations, we can relate the wavelengths:
\[ \lambda_{He^+} = \lambda_H \cdot \frac{1}{z^2} \]

### Step 4: Substitute the known values
Given that the wavelength of the high energy transition of H is \(91.2 \, \text{nm}\), we can substitute:
- \(\lambda_H = 91.2 \, \text{nm}\)
- \(z = 2\)

Now substituting these values into the equation:
\[ \lambda_{He^+} = 91.2 \, \text{nm} \cdot \frac{1}{2^2} = 91.2 \, \text{nm} \cdot \frac{1}{4} \]

### Step 5: Calculate the wavelength for He+
\[ \lambda_{He^+} = 91.2 \, \text{nm} \cdot 0.25 = 22.8 \, \text{nm} \]

### Final Answer
The corresponding wavelength of the He atom (He+) is \(22.8 \, \text{nm}\).

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