Question

In which series of hydrogen spectrum does this line fall?

Hint:

The Balmer series corresponds to transitions where the electron moves to the \( n = 2 \) energy level from higher levels. The lines of the Balmer series fall in the visible spectrum.

Answer 1

Step 1: The wavelength of the emitted radiation corresponds to a specific transition in the hydrogen atom. Based on the energy difference between the levels, we can identify the series of the hydrogen spectrum. Step 2: The energy levels of the hydrogen atom are given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] where \( n \) is the principal quantum number. Step 3: The energy difference between two levels, say \( n_1 \) and \( n_2 \), is given by: \[ \Delta E = E_{n_1} - E_{n_2} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \, \text{eV} \] Step 4: The transition that corresponds to the wavelength \( 486 \, \text{nm} \) is associated with the \( n_2 = 3 \) and \( n_1 = 2 \) levels of the hydrogen atom. This is part of the Balmer series, which involves transitions from higher levels (with \( n_2 \geq 3 \)) to \( n_1 = 2 \). Step 5: Since the wavelength 486 nm corresponds to this transition, it falls in the Balmer series of the hydrogen spectrum.