The wavelength of first line of Balmer series in hydrogen atom is 2, the wavelength of first of corresponding double ionized lithium atom is …………

Options

A. `(lamda)/(3)`
B. `(lamda)/(4)`
C. `(lamda)/(9)`
D. `(lamda)/(27)`

Answer

C

Explanation

`(1)/(lamda)=Z^(2)R[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
Here for first line in Balmer series of hydrogen atom
`(1)/(lamda_(H))=Z_(H)^(2)R[(1)/(n_(f)^(2))-(1)/(3^(2))]`
For `L_(i),(1)/(lamda_(Li))=Z_(Li)^(2)R[(1)/(2^(2))-(1)/(3^(2))]`
`:.(lamda_(Li))/(lamda_(H))=(Z_(H)^(2))/(Z_(Li)^(2))=(1)/(9)" "(Z_(Li)=3,Z_(H)=1)`
`:.lamda_(Li)=(lamda_(H))/(9)` but `lamda_(H)=lamda:.lamda_(Li)=(lamda)/(9)`