1.

For balmer series wavelength of first line is `lambda_(1)` and for brackett series wavelength of first line is `lambda_(2)` then `(lambda_(1))/(lambda_(2))` isA. 0.162B. 0.124C. 0.138D. 0.188

Answer» Correct Answer - A
The wavelength of a line in Balmer series is given by
` ( 1 ) /( lamda ) = R ( ( 1 ) / ( 2 ^ 2 ) - ( 1 ) / ( n ^ 2 )) " " ["for " n = 3, 4 , 5 …] `
where, R = Rydberg constant.
For first line, n = 3
` rArr ( 1 ) /( lamda _ 1 ) = R (( 1 )/ ( 4 ) - ( 1 ) /( 9)) rArr lamda _ 1 = ( 36)/ ( 5R ) " " `...(i)
The wavelength of a line in Brackett series is given by
` ( 1 ) /( lamda ) = R ( ( 1 ) /( 4 ^ 2 ) - ( 1 ) /( n^ 2 )) " "["for " n = 5, 6 , 7...] `
For first line, n = 5
` rArr ( 1 ) / ( lamda _ 2 ) = R ( ( 1 ) /( 16 ) - ( 1 ) /( 25 ) ) rArr lamda _ 2 = (400 ) / ( 9R ) " " `...(ii)
Dividing Eq. (i) by Eq. (ii), we get
` ( lamda _ 1 ) /( lamda _ 2 ) = ( 36 ) /( 5R ) xx ( 9 R ) /( 400) = ( 81 ) /( 500) = 0.162 `


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