Q 12-12-044JEE MainJEE Main 2024 (5 Apr, Shift 2)Medium
The shortest wavelength of the spectral lines in the Lyman series of the hydrogen spectrum is $915\ \text{\AA}$. The longest wavelength of the spectral lines in the Balmer series will be ______ Å.
Numerical value type. Enter your answer.
Answer: 6588
Lyman series limit: $\dfrac{1}{915} = R$.
Longest Balmer line ($3 \to 2$): $\dfrac1\lambda = R\left(\dfrac14 - \dfrac19\right) = \dfrac{5R}{36}$.
$$\lambda = \frac{36}{5}\times915 = 6588\ \text{\AA}$$
Solution by Sreeraj P, M.Sc Physics