Question

Match List I with List II.

List I (Spectral Lines of Hydrogen for transitions from)		List II (Wavelength (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:

• A. A-II, B-I, C-IV, D-III
• B. A-III, B-IV, C-II, D-I
• C. A-IV, B-III, C-I, D-II
• D. A-I, B-II, C-III, D-IV

Answer 1

The Correct Option is B

Step 1: Recall the Balmer Series Formula

The Balmer series describes transitions to \( n_1 = 2 \) in the hydrogen atom. The wavelength of emitted light is given by:

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

Where:

\(\lambda\) = Wavelength of emitted light

R = Rydberg constant

n₁ = Lower energy level (\( n_1 = 2 \) for the Balmer series)

n₂ = Higher energy level (\( n_2 > 2 \))

Step 2: Match Wavelengths

Using known values of wavelengths for each transition:

\( n_2 = 3 \rightarrow n_1 = 2 \) : 656.3 nm → (A-III)

\( n_2 = 4 \rightarrow n_1 = 2 \) : 486.1 nm → (B-IV)

\( n_2 = 5 \rightarrow n_1 = 2 \) : 434.1 nm → (C-II)

\( n_2 = 6 \rightarrow n_1 = 2 \) : 410.2 nm → (D-I)

Step 3: Conclusion

The correct matching is:

A-III

B-IV

C-II

D-I