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If a hydrogen atom emit a photon of ener...

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

`1.05xx10^(-34)J-s`

B

`2.11 xx10^(-34)J-s`

C

`3.16 xx 10^(-34)J-s`

D

`4.22 xx 10^(-34)J-s`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta E =-13.6eV ((1)/(n_(1)^(2))-(1)/(n_(2)^(n))),Delta L =(h)/(2pi)(n_1 -n_(2))`
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Knowledge Check

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

    A
    `3h/pi`
    B
    `2h/pi`
    C
    `h/pi`
    D
    `4h/pi`

  • 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)`

  • An isolated hydrogen atom emits a photon of energy 9 eV. Find momentum of the photons

    A
    `4.8xx 10^(-23)kg-m//s`
    B
    `4.8xx 10^(-27)kg-m//s`
    C
    `4.8xx 10^(-30)kg-m//s`
    D
    `7.8xx 10^(-27)kg-m//s`

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