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
These are Balmer lines. $\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{n_2^2}\right)$, so a larger $n_2$ means a larger energy gap and a shorter wavelength.
The order of wavelengths is therefore $3 \to 2 > 4 \to 2 > 5 \to 2 > 6 \to 2$:
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)
A-III, B-IV, C-II, D-I.
Solution by Sreeraj P, M.Sc Physics