What is the wave number of the lowest energy | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.The lowest energy transition in the Lyman series corresponds to the

Vedclass · Accepted Answer

$\bar{v}=R_{H}\left(\frac{3}{4}\right)$

Vedclass · Answer

(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.The lowest energy transition in the Lyman series corresponds to the transition from the nearest higher energy level,which is $n_2 = 2$.The Rydberg formula for the wave number $\bar{v}$ is given by $\bar{v} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Substituting the values $n_1 = 1$ and $n_2 = 2$:$\bar{v} = R_{H} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$$\bar{v} = R_{H} \left( 1 - \frac{1}{4} \right) \ cm^{-1}$$\bar{v} = R_{H} \left( \frac{3}{4} \right) \ cm^{-1}$

What is the wave number of the lowest energy transition associated with the Lyman series?

Similar Questions

Which one of the following statements is not correct?

When an electron with energy $10.8 \,eV$ collides with a hydrogen $(H)$ atom,what will happen?

Bohr's radius can have

Describe the hydrogen spectrum.

The wavenumber of the first line $(n_2=3)$ in the Balmer series of hydrogen is $\bar{\nu}_1 \ cm^{-1}$. What is the wavenumber (in $cm^{-1}$) of the second line $(n_2=4)$ in the Balmer series of $He^{+}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module