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If the series limit wavelength of the Ly...

If the series limit wavelength of the Lyman series for hydrogen atom is `912 Å`, then the series limit wavelength for the Balmer series for the hydrogen atom is

A

`912 Å`

B

`912 xx 2 Å`

C

`912 xx 4 Å`

D

`(912 )/( 2) Å`

Text Solution

Verified by Experts

The correct Answer is:
C
`1/ lamda = R [ (1)/( n_(1)^(2))-(1)/(n_(2)^(2))]`
for limiting wavelength of Lyman series
` n_1 =1,n_2 =oo (1)/( lamda)=R`
for limiting wavelength of Balmer series
`n_1 =2, n_2 =oo`
` (1)/(lamda_B) =R ((1)/(4)) implies lamda _B = 4/R`
` therefore lamda_B =4 lamda_L = 4 xx 9 12 Å`
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If the series limit of wavelength of the Lyman series for the hydrogen atoms is 912 Å, then the series limit of wavelength for the Balmer series of the hydrogen atom is:

Knowledge Check

  • The series limit wavelength of the Balmer series for the hydrogen atom is

    A
    `1/R`
    B
    `4/R`
    C
    `9/R`
    D
    `16/R`

  • If the series limit wavelength of the Lyman series of hydrogen atom is 912Å , then the series limit wavelength of the Balmer series of the hydrogen atom is

    A
    `912Å`
    B
    `1824Å`
    C
    `3648Å`
    D
    `456Å`

  • If the series limit of wavelength of the Lyman series for the hydrogen atoms is 912 Å, then the series limit of wavelength for the Balmer series of the hydrogen atom is:

    A
    912 Å
    B
    `912 xx 2` Å
    C
    `912xx4` Å
    D
    `912//2` Å

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