The atomic spectrum of hydrogen is found to contain a series of lines of wavelengths 656.46, 486.27, 434.17 and 410.29 nm. The wavelength (in nm) of the next line in the series will be....

The atomic spectrum of hydrogen is found to contain a series of lines

1.	The atomic spectrum of hydrogen is found to contain a series of lines of wavelengths 656.46, 486.27, 434.17 and 410.29 nm. The wavelength (in nm) of the next line in the series will be....
Answer» Solution :The given wavelengths lie in the visible region. HENCE, they are expected to belong to Balmer series. Thus, `n_(1) = 2`. LET us calculate `n_(2)` for the shortest wavelength, viz., 410.29 nm `(1)/(lamda) = R_(H) ((1)/(2^(2)) - (1)/(n_(2)^(2)))` `(1)/(410.29 xx 10^(-7) cm) = 109,67 cm^(-1) ((1)/(4) - (1)/(n_(2)^(2)))` This on solving gives `n_(2) = 6` Thus, the NEXT LINE will be OBTAINED for jump from `n_(2) = 7 " to " n_(1) = 2`, so that `(1)/(lamda) = 109,677 cm^(-1) ((1)/(2^(2)) - (1)/(7^(2)))` `= 109,677 xx ((1)/(4) - (1)/(49)) cm^(-1)` `= 109,677 xx (45)/(196) cm^(-1) = 25180.9 cm^(-1)` or `lamda = (1)/(25180.9 cm^(-1)) = 397.1 xx 10^(-7) cm` `= 397.13 nm`