The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio

Step-by-step solution

Wavelength of spectral lines in H -atom is given as 1 =R ( 1 n₁² - 1 n₂² ) ...(i) For shortest wavelength of Paschen series, n₁=3 and n₂= From Eq. (i), we get aligned & l _ P S =R ( 1 3² - 1 )= R 9 & _ P S = 9 R ...(ii) aligned For shortest wavelength in Balmer series, n₁=2, n₂= From Eq. (i), we get 1 _ B S =R ( 1 2² - 1 )= R 4 _ B S = 4 R ...(iii) For shortest wavelength of Lyman series, n₁=1 and n₂= From Eq. (i), we get aligned 1 _ L S &=R ( 1 1² - 1 )=R _ L S &= 1 R aligned Hence, from Eq. (i), (ii) and (iii), w