The wavelength of the spectral live the Balner series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectralline in the Balmer series of singly - ionized belium atom is
A
`1215 A^(2) `
B
`1640 A^(2) `
C
`2430 A^(2) `
D
`4687 A^(2) `
Text Solution
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The correct Answer is:
A
we know that `(1)/(lambda) = RZ^(2) [(1)/(n_(1)^(2)) - (1)./(n_(2)^(2))] The wave length of spectal line in the balmar series of hydrogen atom is `6561Å` Here n_(2) = 3 and n_(1) = 2 ` ` :. (1)/(6561) = R(1)^(2) ((1)/(4) - (1)/(9)) = (5R)/(36) ` ....(i) For the second spectral line is the balmer of singly ionised belium ion `n_(2) = 4 and n_(1) = 2 , Z = 2 ` `:. (1)/(lambda) = R(2)^(2) [(1)/(4) - (1)/(16)] = (3R)/(4) `....(ii) Dividingh equation (i) and equqtion (ii() we get `(lambda)./(6561) = (5R)/(36) xx (4)/(3R) = (5)/(27)` ` :,. lambda= 1215 Å`
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