Question

From Lyman series of Pfund series, frequency___________
A. Increase
B. Decrease
C. Increase or decrease
D. No change

Answer 1

Hint: There are five kinds of series in the hydrogen spectrum. Each series gives ideas about wave numbers. Whereas reciprocal of wave number is wavelength. Wave number gives the number of lines in unit distance. Formula used in the hydrogen spectrum is the Bohr formula.

Complete step by step answer:
You must have learned about the hydrogen spectrum. In the hydrogen spectrum we have a number of series. Now will see how frequency will vary with series but before this let’s see some information about this series.
Lyman series: This series arises due to transition of an electron from different outer orbits to the first bohr orbit. Therefore the wavelength for Lyman series is given by
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{n}^{2}}} \right);where,n=2,3,4,5....$
For n= 2 and 3; the wavelengths are 1216 ${{A}^{0}}$ and 1025 ${{A}^{0}}$ respectively. As the value of n increases wavelength decreases. It means frequency will increase.
Balmer series: This series original is due to the transition of the electrons from different outer Orbit to the second Orbit. Therefore the wavelength of the Balmer series is given by,
 $\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{n}^{2}}} \right);where,n=3,4,5....$
For n= 3 and 4 the well and are 6563${{A}^{0}}$ and 4868${{A}^{0}}$ respectively. As the value of n increases, wavelength decreases. It means frequency will increase.
Paschen series: A series of spectral lines originates from the transition from different outer space to the third orbit. Wavelength of Paschen series is given by
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{n}^{2}}} \right);where,n=4,5,6,7....$
For n= 4 and 5, the wavelengths are 18750${{A}^{0}}$and 18750${{A}^{0}}$ respectively.
Brackett series: This series arises due to transition of the electron from the different outer orbits to the fourth orbit to the fourth orbit. The wavelength of brackett series is given by,
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{n}^{2}}} \right);where,n=5,6,7....$
For n= 5,6. The wavelengths are 40518 and 26253 respectively. As the value of n increases wavelength decreases. It means frequency will increase.
Pfund series: This series arises due to the transition of electrons from different outer Orbit to fifth orbit. The wavelength of Pfund series is given by,
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{5}^{2}}}-\dfrac{1}{{{n}^{2}}} \right);where,n=6,7....$
For n=6 and 7: the wavelengths are 75587${{A}^{0}}$ and 46533${{A}^{0}}$ respectively. As the value of n increases, wavelength decreases. It means frequency will increase.

So as you can see, starting from Lyman to Pfund, wavelengths continuously increase. Where frequency is inversely proportional to wavelength. Therefore if wavelength is increased then frequency must be decreased.

Hence, the correct answer is option B.

Note:
The Lyman series lies in the ultraviolet region of the spectrum. The Balmer series lies in the visible region of the Spectrum. The Paschen series lies in the infrared region of the spectrum. The Brackett series lies near the infrared region of the spectrum. The Pfund series lies in the far infrared region of the spectrum.