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Grade: 11
ratio between longest wavelengths of H atom in lyman series to the shortest wavelengths wavelength in balmer series of He+ is?
2 years ago

Answers : (2)

114 Points
							Wave no. = R [1/n1^2 - 1/n2^2] = 1/wavelengthTo find ratio of longest wavelength of Him Lyman and shortest of He+ in balmer series just find ratio of wave no. Of He+ to the H= R [1/2^2 - 1/~]/R [1/1^2 - 1/2^2]~= infinityLongest wave length when n2=n1+1And shortest when n2=~So ratio comes out to be 1:3
2 years ago
Mayur Jatinbhai Brahmbhatt
13 Points
1/LAMDA = RZ^2(1/22 - 1/32)
                  = R×4×(5/36)
                  = 5/9R
So the ratio is 5/9
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one year ago
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