Question

A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are their wavelengths?

Answer 1

Answer:

$\lambda=25.6nm$

Explanation:

The Rydberg formula can be extended for use with any hydrogen-like chemical elements, that is to say with only one electron being affected by effective nuclear charge. So, in this case, we can calculate the wavelenghts of the emitted photons using this formula:

$\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where R is the Rydberg constant of the element, Z its atomic number, $n_1$ is the lower energy level and $n_2$ the upper energy level of the  electron transition. Recall that the ground state is denoted as n=1.

$\frac{1}{\lambda}=1.1*10^7m^{-1}(2)2^2(\frac{1}{1^2}-\frac{1}{3^2})\\\frac{1}{\lambda}=3.91*10^7m^{-1}\\\lambda=2.56*10^{-8}m=25.6nm$