Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie?
Solution:
Wavelength in the Balmer series of H-atom: 1λ=1.097×107×(1n12−1n22)m−1 For shortest wavelength in the Balmer series: n1=2 n2=∞ ∴1λmin=1.097×107×(122−1∞)m−1 λmin=41.097×107m=3.646×10−7m=364.6nm This wavelength lies in the ultraviolet region.