Respuesta :
Answer and Explanation:
The computation of the shortest wavelength in the series is shown below:-
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
Where
[tex]\lambda[/tex] represents wavelength
R represents Rydberg's constant
[tex]n_f[/tex] represents Final energy states
and [tex]n_i[/tex] represents initial energy states
Now Substitute is
[tex]1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
now we will put the values into the above formula
[tex]= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )[/tex]
[tex]= 1218888.889 m^{-1}[/tex]
Now we will rewrite the answer in the term of [tex]\lambda[/tex]
[tex]\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m[/tex]
So, the whole Paschen series is in the part of the spectrum.
