If the shortest wavelength of the Lyman series of an H atom is , then the wavelength of the 1st line — Atomic Structure Chemistry Question

💡 Solution & Explanation

Shortest Lyman line ($n=\infty \to n=1$): $1/x = R(1/1 - 0) = R$. 1st Balmer line ($n=3 \to n=2$): $1/\lambda = R(1/4 - 1/9) = 5R/36$. Substituting $R = 1/x$, we get $1/\lambda = 5/(36x)$, thus $\lambda = 36x/5$.