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Taking the wavelength of first Balmer li...

Taking the wavelength of first Balmer line in hydrogen spectrum (`n = 3` to `n = 2`) as 660 nm, the wavelength of the `2^(nd)` Balmer line (`n = 4` to `n = 2`) will be:

A

889.2 nm

B

488.9 nm

C

388.9 nm

D

642.7 nm

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • Taking the wavelength of first Balmer line in the hydrogen spectrum (n=3" to "n=2) as 660 nm, then the wavelength of 2^("nd") Balmer line in the same spectrum (n=4" to "n=2) will be

    A
    488.9 nm
    B
    388.9 nm
    C
    889.2 nm
    D
    642.7 nm

  • If the wavelength of the first line of the Balmer series of hydrogen atom is 656.1 nm the wavelngth of the second line of this series would be

    A
    `218.7 nm`
    B
    `328.0 nm`
    C
    `486.nm`
    D
    `640.0nm`

  • If the wavelength of the first line of the Balmer series of hydrogen is 6561 A, the wavelength of the second line of the series should be.

    A
    `13122 Å`
    B
    `3280Å`
    C
    `4860Å`
    D
    `2178Å`

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