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).
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).
Answer:
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]