Q 12-12-005NEETNEET 2024Top questionMedium
Match List I with List II.
List I (Spectral Lines of Hydrogen for transitions from):
A. $n_2 = 3$ to $n_1 = 2$
B. $n_2 = 4$ to $n_1 = 2$
C. $n_2 = 5$ to $n_1 = 2$
D. $n_2 = 6$ to $n_1 = 2$
List II (Wavelengths (nm)):
I. $410.2$
II. $434.1$
III. $656.3$
IV. $486.1$
Choose the correct answer from the options given below :
Answer: (D) A-III, B-IV, C-II, D-I
Concept. All four are Balmer lines: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{n_2^2}\right)$. A larger $n_2$ gives a larger energy gap and a shorter wavelength.
Step 1. Order the wavelengths. $\lambda_{3\to2} > \lambda_{4\to2} > \lambda_{5\to2} > \lambda_{6\to2}$.
Step 2. Match.
- A ($3 \to 2$, H$_\alpha$): $656.3$ nm → III.
- B ($4 \to 2$, H$_\beta$): $486.1$ nm → IV.
- C ($5 \to 2$, H$_\gamma$): $434.1$ nm → II.
- D ($6 \to 2$, H$_\delta$): $410.2$ nm → I.
Answer. A-III, B-IV, C-II, D-I (Option D).
Solution by Sreeraj P, M.Sc Physics