Question

question 12. deduce the wavelength of the electromagnetic radiation emitted when an electron transitions from a higher energy level to the second main energy level (n = 2).

Explanation:

Step1: Recall energy - wavelength relationship

The energy of electromagnetic radiation is given by $E = h
u=\frac{hc}{\lambda}$, where $h$ is Planck's constant, $
u$ is the frequency, $c$ is the speed of light and $\lambda$ is the wavelength. When an electron makes a transition from a higher energy level to $n = 2$, the energy of the emitted photon is related to the wavelength. Transitions to $n=2$ give rise to the Balmer series. The visible - light region corresponds to Balmer series transitions.

Step2: Identify the answer

Since the transition is to $n = 2$ and the question asks for the wavelength of the electromagnetic radiation, and transitions to $n = 2$ in hydrogen - like atoms emit visible light, the electromagnetic radiation is visible light.

Answer:

Visible light