Question

Which series has the highest energy in the hydrogen spectrum?
a.) Balmer
b.) Brackett
c.) Pfund
d.) Lyman

Answer 1

Hint: Energy can be calculated as well as compared using Rydberg's formula which is used to find out wavelength and also for Lyman, n = 1, and for Balmer, n = 2. We will use this information to find out the answer.

Complete step by step answer:
- These names Lyman, Balmer, Brackett, Pfund are given to the lines in hydrogen atom spectra. If we look at the spectra of hydrogen atoms, we find that the emission spectra are divided into a number of lines which are formed due to electron making transitions between respective energy levels which result in the formation of lines.
- Lyman series: when an electron makes a transition from an outer orbit to the first orbit i.e., n=1, that line is called the Lyman series.
- Balmer series: when an electron makes a transition from an outer orbit to a second orbit i.e., n=2, that line is called the Balmer series.
- Similarly, for Brackett n = 4 and Pfund n = 5.
- Now, calculate the energy using the formula shown below
\begin{align*}
\Delta E = 13.6(\dfrac{1}{n^{2}_{1}} -\dfrac{1}{n^{2}_{2}})
\end{align*}
For Lyman,
\begin{align*}n_{1} = 1,\ n_{2} = 2\end{align*}
So, \begin{align*}\Delta E=10.2eV\end{align*}
For Balmer,
\begin{align*}n_{1}=2,\ n_{2} = 3\end{align*}
So, \begin{align*}\Delta E = 1.88 eV\end{align*}
- Now, as we can observe, by moving towards the end of the spectra, energy reduces.
So, the correct answer is “Option D”.

Note: This equation gets modified for Hydrogen atoms and general questions are asked from this equation but for other atoms, the Rydberg equation will be used and few properties like dependence of wavelength, atomic number, etc. should be known from that equation.