For helium, let’s label the emission wavelengths, λ(nf,ni) where nf and ni are, respectively, the principal quantum numbers of the final and

Question

For helium, let’s label the emission wavelengths, λ(nf,ni) where nf and ni are, respectively, the principal quantum numbers of the final and the initial states. Calculate the following emission wavelengths: λ(3,4), λ(4,6), λ(4,7) λ(4,8) and λ(4,9).

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Hải Đăng 1 year 2021-08-31T15:03:29+00:00 1 Answers 10 views 0

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  1. Adela
    0
    2021-08-31T15:04:45+00:00
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    Answer:

    • The emission wavelength for  λ(3,4) is 1875.24 nm
    • The emission wavelength for  λ(4,6) is 2625.34 nm
    • The emission wavelength for  λ(4,7) is 2165.69 nm
    • The emission wavelength for  λ(4,8) is 1944.70 nm
    • The emission wavelength for  λ(4,9) is 1817.54 nm

    Explanation:

    Using Rydberg equation

    [tex]\frac{1}{\lambda} = -R(\frac{1}{n_f^2} – \frac{1}{n_i^2}) = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})[/tex]

    where;

    R is Rydberg equation = 1.097 x 10⁷ m⁻¹

    For λ(3,4)

    [tex]\frac{1}{\lambda} = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{3^2} – \frac{1}{4^2}) = 533263.889 \ m^{-1}\\\\\lambda = 1875.24 \ nm[/tex]

    For λ(4,6)

    [tex]\frac{1}{\lambda} = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{4^2} – \frac{1}{6^2}) = 380902.778 \ m^{-1}\\\\\lambda = 2625.34 \ nm[/tex]

    For λ(4,7)

    [tex]\frac{1}{\lambda} = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{4^2} – \frac{1}{7^2}) = 3461747.45 \ m^{-1}\\\\\lambda = 2165.69 \ nm[/tex]

    For  λ(4,8)

    [tex]\frac{1}{\lambda} = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{4^2} – \frac{1}{8^2}) = 514218.75 \ m^{-1}\\\\\lambda = 1944.70 \ nm[/tex]

    For  λ(4,9)

    [tex]\frac{1}{\lambda} = R(\frac{1}{n_i^2} – \frac{1}{n_f^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{4^2} – \frac{1}{9^2}) = 550192.90 \ m^{-1}\\\\\lambda = 1817.54 \ nm[/tex]

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