Answer :
Answer:
[tex]n=4[/tex]
Explanation:
Paschen series of hydrogen, is the series of transitions resulting from the hydrogen atom when electron jumps from a state of [tex]n\geq 4[/tex] to [tex]n_f=3[/tex].
Emission lines for hydrogen are given by:
[tex]\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n^2})[/tex]
[tex]\lambda[/tex] is the wavelength of the line emitted,[tex]R_H[/tex] is the Rydberg constant for hydrogen, [tex]n_f[/tex] is the final energy state of the electron and [tex]n[/tex] is the energy state where the electron transition originated.
We have [tex]\lambda=1880nm=1880*10^{-9}m[/tex] and [tex]n_f=3[/tex]. Solving for n:
[tex]\frac{1}{\lambda R_H}=\frac{1}{n_f^2}-\frac{1}{n^2}\\\frac{1}{n^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R_H}\\n^2=\frac{1}{\frac{1}{n_f^2}-\frac{1}{\lambda R_H}}\\n^2=\frac{1}{\frac{1}{3^2}-\frac{1}{1880*10^{-9}m(1.09737*10^7m^{-1})}}\\n^2=15.96\\n=4[/tex]