Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

Advertise your business, and reach millions of students around the world.

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

Q. 1.3

Calculate the wavelength of the Balmer series limit.

Verified Solution

When an electron jumps from outer orbits to the second orbit, the series is called Balmer series. Using Eqn. (1.8), the wavelength limit for the Balmer series can be found by calculating the wavelength of the radiation due to the transition of electron from the infinite orbit to the second orbit.

$\lambda =\frac{12,400}{E_{2}-E_{1} }$   (1.8)

Wavelength of the Balmer series limit      $=\frac{12,400}{E_{\infty }-E_{2} }$

Energy of the electron at the infinite orbit, $E_{\infty }=\frac{ -13.6 }{\infty ^{2}}=0$

Energy of the electron at the second orbit, $E_{2}=\frac{-13.6}{2^{2} }=-3.4$

Therefore, the wavelength limit         $=\frac{12,400}{E_{\infty }-E_{2}} =\frac{12,400}{3.4}=3647A.U.$