Question

When an electron does transition from $n = 4$ to $n = 2$, then emitted line spectrum will be

• A. first line of Lyman series
• B. second line of Balmer series
• C. first line of Paschen series
• D. second line of Paschen series.

Answer 1

The Correct Option is B

To determine the emitted line spectrum when an electron transitions from n = 4 to n = 2 in a hydrogen atom, we need to understand the Balmer series. The Balmer series is a part of the emission spectrum of hydrogen when the electron transitions to the second energy level (n = 2) from a higher energy level.

The general formula for the wavelength of light emitted in the Balmer series is given by the Rydberg formula:

\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

where R is the Rydberg constant, n_1 is the final energy level (for the Balmer series, n_1 = 2), and n_2 is the initial energy level.

In this case, the transition is from n = 4 to n = 2. Thus, we can substitute n_1 = 2 and n_2 = 4 into the formula:

\frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{4^2}\right)

\frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{16}\right)

\frac{1}{\lambda} = R \left(\frac{4 - 1}{16}\right) = R \left(\frac{3}{16}\right)

This transition corresponds to the second line of the Balmer series, as the first line occurs when an electron transitions from n = 3 to n = 2. The Balmer series is characterized by transitions to n_1 = 2, and the correct answer is the second line of Balmer series.

Thus, the correct option is: second line of Balmer series.