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In the spectrum of singly ionized helium...

In the spectrum of singly ionized helium , the wavelength of a line abserved is almost the same as the first line of Balmer series of hydrogen . It is due to transition of electron from `n_(1) = 6 to n_(2) = ^(''**'')`. What is the value of `(''**'')`.

Text Solution

Verified by Experts

The correct Answer is:
4
For the first line of Balmer series of hydrogen
`(1)/(lambda) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36) implies lambda = (36)/(5 R)`
"For singly ionized helium" `(Z = 2)`,
`(1)/(lambda) = 4 R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
Given `lambda = lambda implies 4 R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (5R)/(36)`
`implies n_(2) = 4`
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Knowledge Check

  • The third line of the Balmer series, in the emission spectrum of the hydrogen atom, is due to the transition from the

    A
    fourth Bohr orbit to the first Bohr orbit
    B
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    C
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    D
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  • Wavelength of first line in Lymen series is lambda . What is wavelength of first line in Balmer series?

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    D
    `(5 lambda)/36`

  • The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

    A
    912 Å
    B
    1215 Å
    C
    1512 Å
    D
    6563 Å

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