Wave length of particular transition for H atom is 400nm. What can be wavelength of `He^(+)` for same transition:

Options

A. 400nm
B. 100nm
C. 1600nm
D. 200nm

Explanation

To solve the problem of finding the wavelength of a transition in the He\(^+\) ion given that the wavelength for the same transition in the hydrogen atom (H) is 400 nm, 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 light emitted during a transition between two energy levels in a hydrogen-like atom 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 light,
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(Z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the two energy levels (with \(n_2 > n_1\)).

### Step 2: Apply the Formula for Hydrogen (H)
For hydrogen (\(Z = 1\)), the equation becomes:

\[ \frac{1}{\lambda_1} = R \cdot 1^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Given that \(\lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\), we can express this as:

\[ \frac{1}{400 \times 10^{-9}} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 3: Apply the Formula for Helium Ion (He\(^+\))
For the helium ion (\(Z = 2\)), the equation becomes:

\[ \frac{1}{\lambda_2} = R \cdot 2^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda_2} = 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Relate the Two Equations
Now, we can relate the two equations:

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
\[ \frac{1}{\lambda_2} = 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Dividing the second equation by the first:

\[ \frac{1/\lambda_2}{1/\lambda_1} = \frac{4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)}{R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)} \]

This simplifies to:

\[ \frac{\lambda_1}{\lambda_2} = 4 \]

### Step 5: Solve for \(\lambda_2\)
Rearranging gives:

\[ \lambda_2 = \frac{\lambda_1}{4} \]

Substituting \(\lambda_1 = 400 \, \text{nm}\):

\[ \lambda_2 = \frac{400 \, \text{nm}}{4} = 100 \, \text{nm} \]

### Final Answer
Thus, the wavelength of the transition for He\(^+\) is:

\[ \lambda_2 = 100 \, \text{nm} \]