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In the spectrum of hydrogen atom, the ra...

In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:

A

`(5)/(27)`

B

`(4)/(9)`

C

`(9)/(4)`

D

`(27)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(lamda)=Rz^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
Lyman series: `n_(1)=1.n_(2)=2`
`(1)/(lamda_(1))=Rz^(2)((1)/(1^(2))-(1)/(2^(2)))` . . . (1)
Balmer series:
`n_(1)=2,n_(2)=3`
`(1)/(lamda_(2))=Rz^(2)((1)/(2^(2))-(1)/(3^(2)))` . .. (2)
`(eq^(n)(2))/(eq^(n)(1))implies(lamda_(1))/(lamda_(2))=(Rz^(2)((1)/(4)-(1)/(9)))/(Rz^(2)(1-(1)/(4)))=((5)/(36))/((3)/(4))`
`=(5)/(36)xx(4)/(3)=(5)/(27)`
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