Bohr Model of the Hydrogen atom.
Question

# Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

Moderate
Solution

## The wavelength of different spectral lines of Lyman series is given by$\frac{1}{{\lambda }_{L}}=R\left[\frac{1}{{l}^{2}}-\frac{1}{{n}^{2}}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}where\text{\hspace{0.17em}\hspace{0.17em}}n=2,3,4,....$where subscript L refers to Lyman. For longest wavelength, n = 2$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\lambda }_{{L}_{longest}}}=R\left[\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right]=\frac{3}{4}R\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}---\left(i\right)$The wavelength of different spectral series of Balmer series is given by$\frac{1}{{\lambda }_{B}}=R\left[\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right]\text{\hspace{0.17em}\hspace{0.17em}}where\text{\hspace{0.17em}\hspace{0.17em}}n=3,4,5,....$where subscript B refers to Balmer. For longest wavelength, n= 3$\therefore \text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\lambda }_{{B}_{longest}}}=R\left[\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5R}{36}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}---\left(ii\right)$Divide (ii) by (i), we get$\frac{{\lambda }_{{L}_{longest}}}{{\lambda }_{{B}_{longest}}}=\frac{5R}{36}×\frac{4}{3R}=\frac{5}{27}$

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