For Li2+ ion, calculate,wavelength of the light that is emitted upon the electron's transition from n=4 to n=3?

3 Answers

  • mRNA
    Lv 5
    1 decade ago
    Favorite Answer

    As Li2+ is a hydrogen-like ion, you can use the Rydberg formula

    ΔE = RchZ²{1/n²(1) - 1/n²(2)}

    where R is the Rydberg constant (1.0974 x 10E7 m-¹); c the speed of light in vacuum (2.9979 x 10E8 m s-¹); h Planck's constant (6.626 x 10E-34 J s); Z nuclear charge; n(1) & n(2) are the principal quantum numbers of the electron transition.

    In this case Z = 3; n(1) = 3; n(2) = 4.

    Substituting gives energy difference, ΔE = 9.514 x 10E-19 J (to 4 s.f.)

    The Planck-Einstein relation states ΔE = hf; where f is the frequency of EM radiation emitted or absorbed to give rise to an electronic transition.

    For waves, f = c/λ; where λ is wavelength.

    So ΔE = hc/λ. Hence λ = 2.088 x 10E-7 m = 208.8 nm (to 4 s.f.)

  • Anonymous
    4 years ago

    What Is Emitted

  • 6 years ago

    if the question states is doubly charge then we multiply the wavelength by two?

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