Question

Among the first lines Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has higher energy?
A. Lyman
B. Balmer
C. Paschen
D. Brackett

Answer 1

Energy can be calculated as well as compared using the Rydberg’s formula which is used to find out wavelength and also For Lyman $n{\text{ }} = 1$, for Balmer$n = 2$. This much information is sufficient enough to answer the question.

Step by step explanation:
These names Lyman, Balmer, Paschen, Brackett 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 results in formation of lines.
Lyman series: - When an electron makes a transition from any outer orbit to first orbit i.e. $n = 1$, that line is called a Lyman series.
Balmer Series: - When an electron makes transition from any outer orbit to second orbit i.e. $n = 2$then the line corresponding to it is called the Balmer series.
Similar for Paschen $n = 3$, for Brackett $n{\text{ }} = {\text{ }}4.$
Now to calculate formula called Rydberg’s formula is used which is as follows: -
$\dfrac{1}{\lambda } = {R_h}{Z^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{n_2^2}}} \right]$ where Rh = Rydberg constant, Z = effective nuclear charge, n = energy level.
Also, we know that $E = h\nu = h\dfrac{c}{\lambda }$
So, by putting constant value and other values for Hydrogen atom we get
$\Delta E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
For the first line i.e. Lyman ${n_1} = {\text{ }}1,{\text{ }}{n_2} = {\text{ }}2$
So, $\Delta {\text{ }}E{\text{ }} = {\text{ }}10.2eV$
For Balmer ${n_1}{\text{ }} = {\text{ }}2{\text{ }}and{\text{ }}{n_2}{\text{ }} = {\text{ }}3$
$\Delta E = 1.88eV$
So, we see that as we are moving towards end of the spectra, energy is decreasing

Therefore, option A is the correct answer i.e. Lyman which has higher energy among all of these.

Note: This equation gets modified for Hydrogen atom 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.