Question

A line in the paschen series of hydrogen has a wavelength of 1880 nm. From what state did the electron originate?

Answer 1

Answer:

$n=4$

Explanation:

Paschen series of hydrogen, is the series of transitions resulting from the hydrogen atom when electron jumps from a state of $n\geq 4$ to $n_f=3$.

Emission lines for hydrogen are given by:

$\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n^2})$

$\lambda$ is the wavelength of the line emitted,$R_H$ is the Rydberg constant for hydrogen, $n_f$ is the final energy state of the electron and $n$ is the energy state where the electron transition originated.

We have $\lambda=1880nm=1880*10^{-9}m$ and $n_f=3$. Solving for n:

$\frac{1}{\lambda R_H}=\frac{1}{n_f^2}-\frac{1}{n^2}\\\frac{1}{n^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R_H}\\n^2=\frac{1}{\frac{1}{n_f^2}-\frac{1}{\lambda R_H}}\\n^2=\frac{1}{\frac{1}{3^2}-\frac{1}{1880*10^{-9}m(1.09737*10^7m^{-1})}}\\n^2=15.96\\n=4$