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 :

To match each transition in List I with its wavelength in List II we use the Rydberg formula for hydrogen spectral lines:

$1/\lambda \; =\; R\; (1/n₁² - 1/n₂²)$

For the Balmer series the lower level is _n₁=2. The Rydberg constant for hydrogen is $R \approx 1.097 \times 10⁷ m⁻¹$. Substituting _n₁=2 gives

$1/\lambda \; =\; R\; (1/2² - 1/n₂²)\; =\; R\; (1/4 - 1/n₂²)$

Now calculate λ for each upper level _n₂:

Transition A: n₂ = 3 → n₁ = 2

$1/\lambda \; =\; R\; (1/4 - 1/9)\; =\; R\; (5/36)$

λ ≈ 656.3 nm → matches item III.

Transition B: n₂ = 4 → n₁ = 2

$1/\lambda \; =\; R\; (1/4 - 1/16)\; =\; R\; (3/16)$

λ ≈ 486.1 nm → matches item IV.

Transition C: n₂ = 5 → n₁ = 2

$1/\lambda \; =\; R\; (1/4 - 1/25)\; =\; R\; (21/100)$

λ ≈ 434.1 nm → matches item II.

Transition D: n₂ = 6 → n₁ = 2

$1/\lambda \; =\; R\; (1/4 - 1/36)\; =\; R\; (8/36)$

λ ≈ 410.2 nm → matches item I.

Therefore the correct matching is:

A – III,  B – IV,  C – II,  D – I.