The wavelength of Hα line of Balmer series is XÅ. What is the X of Hβ line of Balmer series?
(a) X\(\frac{108}{80}\)
(b) X\(\frac{80}{108}\)
(c) \(\frac{1}{X}\frac{80}{108}\)
(d) \(\frac{1}{X}\frac{108}{80}\)
Correct answer: (b) X\(\frac{80}{108}\)
Explanation:
Hα line of Balmer series means first line of Balmer series.
n1 = 2, n2 = 3
\(\bar{v} = \frac{1}{\lambda_{\alpha}}\) = R(\(\frac{1}{2^2} - \frac{1}{3^2}\)) = \(\frac{5R}{36}\)
∴ \(\lambda_{\alpha} = \frac{36}{5R}\) = X
Hβ line of Balmer series means, second line of Balmer series, n1 = 2, n2 = 4.
\(\bar{v} = \frac{1}{\lambda_{\beta}}\) = R(\(\frac{1}{2^2} - \frac{1}{4^2}\)) = \(\frac{3R}{16}\)
∴ \(\lambda_{\beta} = \frac{16}{3R}\) = X
when \(\frac{36}{3R}\) = X
Then \(\frac{16}{3R}\) = \(\frac{X \times 5R \times 16}{36 \times 3R}\)
= X\(\frac{80}{108}\)