The Lyman series of hydrogen spectrom can be respectively by the equation
`v = 3.28 xx 10^(15)[(1)/(1^(2)) - (1)/(n^(2))] s^(-1)`
Calculate the maximum and minimum frequencies in this series

Answer

A, B, C, D

Explanation

Lyman frequrncy will be maximum corresponding to the maximum energy transition i.e. `1 rarr oo`
`rArr v_(max) = 3.28 xx 10^(15) [(1)/(1^(2)) - (1)/(oo^(2))] s^(-1) = 3.26 xx 10^(15) s^(-1)`
Note that coreresponding wavelength will be shortest wavelength
And Lyman frequency will be coreresponding to minimum energy transition i.e. `1 rarr 2`
`rArr v_(min) = 3.28 xx 10^(15) [(1)/(1^(2)) - (1)/(2^(2))] s^(-1) = 2.46 xx 10^(15) s^(-1)`
Note that coreresponding wavelength will be longest wavelength.