Question

Calculate the shortest and longest wavelengths of Balmer series of hydrogen atom. Given $R=1.097\times {10}^7{m}^{-1}$.

Answer 1

Wavelength of photon emitted due to transition in H-atom $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Shortest wavelength is emitted in Balmer series if the transition of electron takes place from $n_2=\infty$ to $n_1=2$.

$\therefore$ Shortest wavelength in Balmer series $\frac{1}{{\lambda }_s}=R(\frac{1}{2^2}-\frac{1}{\infty })$

Or ${\lambda }_s=\frac{4}{R}$

$⟹$ ${\lambda }_s=\frac{4}{1.097\times {10}^7}=364.6nm$

Longest wavelength is emitted in Balmer series if the transition of electron takes place from $n_2=3$ to $n_1=2$.

$\therefore$ Longest wavelength in Balmer series $\frac{1}{{\lambda }_L}=R(\frac{1}{2^2}-\frac{1}{3^2})$

Or ${\lambda }_s=\frac{36}{5R}$

$⟹$ ${\lambda }_s=\frac{36}{5\times 1.097\times {10}^7}=656.3nm$