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In the spectrum of hydrogen, the ration ...

In the spectrum of hydrogen, the ration of the longest wavelength in the Lyman series to the longest wavelength 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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