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When a hydrogen atoms emits a photon of ...

When a hydrogen atoms emits a photon of energy `12.1 eV` , its orbital angular momentum changes by (where h os Planck's constant)

Text Solution

Verified by Experts

`E_(n_(2)) - E_(n_(1)) = 12.1 eV = 13.6 ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
This equation is correct for transition from `n_(2) = 3 " to " n_(1) = 1`
Angular momentum `= (nh)/(2pi)`
`:.` Change in angular momentum `= (h)/(2pi) (n_(2) - n_(1)) = (h)/(2pi) (3 -1) = (h)/(pi) = (6.626 xx 10^(-34)Js)/(3.14)`
`= 2.11 xx 10^(-14) Js`
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Knowledge Check

  • If a hydrogen atom emit a photon of energy 12.1 eV , its orbital angular momentum changes by Delta L. then Delta L equals

    A
    `1.05 xx 10^(-34) J s`
    B
    `2.11 xx 10^(-34) J s`
    C
    `3.16 xx 10^(-34) J s`
    D
    `4.22 xx 10^(-34)`
  • In hy drogen spectrum, a hydrogen atom emits a photon of wavelength 1027 Å its angular momentum changes by

    A
    `(h)/(pi)`
    B
    `(h)/(2 pi)`
    C
    `(3 h)/(2 pi)`
    D
    `(2 h)/(pi)`
  • When an electron in the hydrogen atom in ground state absorb a photon of energy 12.1eV , its angular momentum

    A
    decrease by `2.11xx10^(-34)`
    B
    decreaseby`1.055xx10^(-34)`
    C
    increaseby `2.11xx10^(-34)`
    D
    increase by `1.055xx10^(-34)`
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