Question

The transition of an electron in hydrogen atom that emits a photon whose wavelength lies in the ultraviolet region of the electromagnetic spectrum is

• A. \( 5 \rightarrow 4 \)
• B. \( 4 \rightarrow 3 \)
• C. \( 3 \rightarrow 2 \)
• D. \( 2 \rightarrow 1 \)

Hint:

Remember, the Balmer series corresponds to transitions in hydrogen that emit visible light, while transitions to the first energy level correspond to ultraviolet radiation.

Answer 1

The Correct Option is D

When an electron in a hydrogen atom transitions from a higher energy level to a lower energy level, it emits a photon. The wavelength of the emitted photon determines which region of the electromagnetic spectrum the transition occurs in.
In the case of the ultraviolet region, the transition that occurs is from the second energy level (n=2) to the first energy level (n=1), which corresponds to the emission of ultraviolet light.
Thus, the transition is \( 2 \rightarrow 1 \).
Therefore, the correct answer is \( \boxed{2 \rightarrow 1} \).

Answer 2

Step 1: Understand hydrogen atom transitions.
In a hydrogen atom, when an electron transitions from a higher energy level to a lower energy level, it emits a photon whose energy corresponds to the difference in energy levels.

Step 2: Spectral series in hydrogen.
Different spectral regions are associated with different electronic transitions:
- Lyman series: transitions to \( n = 1 \) (ultraviolet region)
- Balmer series: transitions to \( n = 2 \) (visible region)
- Paschen series: transitions to \( n = 3 \) (infrared region)

Step 3: Identify transition to ultraviolet region.
Since the ultraviolet region corresponds to the Lyman series, the transition must end at \( n = 1 \).
Among these, the transition from \( n = 2 \rightarrow 1 \) emits the photon with the greatest energy in the Lyman series and lies in the ultraviolet region.

Step 4: Conclusion.
The correct transition that emits a photon in the ultraviolet region is: \( \boxed{2 \rightarrow 1} \)