Match List I with List II.

	List I (Spectral Lines of Hydrogen for transitions from)		List II (Wavelengths (nm))
A.	n2 = 3 to n1 = 2	I.	410.2
B.	n2 = 4 to n1 = 2	II.	434.1
C.	n2 = 5 to n1 = 2	III.	656.3
D.	n2 = 6 to n1 = 2	IV.	486.1

Choose the correct answer from the options given below:

The correct option is A-III, B-IV, C-II, D-I.

To determine the wavelengths of the spectral lines for the transitions of a hydrogen atom, we can use the Rydberg formula for hydrogen:

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

where:
• $\lambda$ is the wavelength of the emitted light,
• $R$ is the Rydberg constant ($R\approx 1.097\times {10}^7{\text{m}}^{-1}$),
• $n_1$ is the lower energy level (here, $n_1=2$ for the Balmer series),
• $n_2$ is the higher energy level ($n_2=3,4,5,6$).

From the relationship, we know that as $n_2$ increases, the energy difference between the levels ($\Delta E$) increases. Since energy is inversely proportional to wavelength ($\Delta E=\frac{hc}{\lambda }$), a larger energy transition corresponds to a shorter wavelength.

Let's map the transitions by sorting them in order of increasing energy (and thus decreasing wavelength):

1. Transition A ($n_2=3\to n_1=2$):
This is the lowest energy transition, which corresponds to the longest wavelength. Comparing the options in List II, the longest wavelength is 656.3 nm.
Therefore, A matches with III.

2. Transition B ($n_2=4\to n_1=2$):
This transition has the second lowest energy, which corresponds to the second longest wavelength, 486.1 nm.
Therefore, B matches with IV.

3. Transition C ($n_2=5\to n_1=2$):
This transition corresponds to the third longest wavelength, 434.1 nm.
Therefore, C matches with II.

4. Transition D ($n_2=6\to n_1=2$):
This is the highest energy transition in the list, which corresponds to the shortest wavelength, 410.2 nm.
Therefore, D matches with I.

Combining all the matches, we get:
A-III, B-IV, C-II, D-I