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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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onyebuchinnaji

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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