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 for a helium ion (He⁺) given that the wavelength for the same transition in a hydrogen atom (H) is 400 nm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Wavelength of transition for H atom, λ_H = 400 nm.

2. **Use the Rydberg Formula:**
The Rydberg formula for the wavelength of emitted 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:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively.

3. **Apply the Formula for Hydrogen (H):**
For hydrogen, \( Z = 1 \). Thus, we can write:
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Plugging in the value of \( \lambda_H \):
\[ \frac{1}{400 \, \text{nm}} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Apply the Formula for Helium Ion (He⁺):**
For helium ion, \( Z = 2 \). Thus, we can write:
\[ \frac{1}{\lambda_{He^+}} = R \cdot 2^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda_{He^+}} = 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

5. **Relate the Two Equations:**
From the equations for hydrogen and helium, we can relate the two wavelengths:
\[ \frac{1}{\lambda_{He^+}} = 4 \cdot \frac{1}{\lambda_H} \]
Therefore:
\[ \lambda_{He^+} = \frac{\lambda_H}{4} \]

6. **Substitute the Value of λ_H:**
Now substituting the value of \( \lambda_H \):
\[ \lambda_{He^+} = \frac{400 \, \text{nm}}{4} = 100 \, \text{nm} \]

### Final Answer:
The wavelength of the transition for He⁺ is **100 nm**.

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