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

The Lyman series of hydrogen spectrom can be respectively by the equat

1.	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» Correct Answer - A::B::C::D 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.

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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

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