The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :
(A)4:1✓ Correct
(B)1:2
(C)1:4
(D)2:1
Answer: (A)
Solution
For the shortest wavelength (series limit) the electron falls from n=∞ to n1, so λ1=Rn121. Balmer (n1=2): λB1=4R⇒λB=R4. Lyman (n1=1): λL1=R⇒λL=R1.
λLλB=1/R4/R=4:1.
JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option