When an electron jumps from higher orbit to the second orbit in `He^(+)` ion,the radiation emitted out will be in `(R=1.09times10^(7)m^(-1))`

Options

A. Ultraviolet region
B. Visible region
C. Infrared region
D. X-ray region

Answer

To solve the problem of determining the radiation emitted when an electron jumps from a higher orbit to the second orbit in the He⁺ ion, we can follow these steps:

### Step 1: Understand the Energy Levels
In a hydrogen-like atom such as He⁺ (helium ion), the energy levels are given by the formula:
\[ E_n = -\frac{Z^2 \cdot R_H}{n^2} \]
where:
- \( Z \) is the atomic number (for He, \( Z = 2 \)),
- \( R_H \) is the Rydberg constant for hydrogen (\( R_H = 1.09 \times 10^7 \, m^{-1} \)),
- \( n \) is the principal quantum number (energy level).

### Step 2: Identify the Transition
The electron is transitioning from a higher energy level (let's say \( n = 3, 4, 5, \) etc.) to the second energy level (\( n = 2 \)). The emitted radiation corresponds to the energy difference between these two levels.

### Step 3: Calculate the Wavelength of Emitted Radiation
The wavelength of the emitted radiation can be calculated using the Rydberg formula for hydrogen-like atoms:
\[ \frac{1}{\lambda} = R \left( \frac{Z^2}{n_1^2} - \frac{Z^2}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( n_1 = 2 \) (final state),
- \( n_2 \) is the initial state (higher orbit),
- \( R = 1.09 \times 10^7 \, m^{-1} \).

### Step 4: Determine the Region of Emission
Since the electron is transitioning to the second energy level, this transition falls under the Balmer series, which corresponds to visible light. The Balmer series includes transitions where the final state is \( n = 2 \).

### Conclusion
The radiation emitted when an electron jumps from a higher orbit to the second orbit in the He⁺ ion will be in the visible region of the electromagnetic spectrum.

### Final Answer
The radiation emitted will be in the **visible region**.

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