Question

The wave number of the spectral line in the emission spectrum of Hydrogen will be equal to $\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$ times the Rydberg's constant if the electron jumps from:

• A. $n=3ton=1$
• B. $n=9ton=1$
• C. $n=10ton=1$
• D. $n=2ton=1$

Answer 1

The correct option is A

$n=3ton=1$

Step 1: Given information

Rydberg's constant $\left(R_H\right)$= $1.097x{10}^7{m}^{-1}$

The wave number $\overline{\nu }$ of the emission spectrum of Hydrogen= $\frac{8}{9}{R}_H$

Step 2: Explanation

1. The hydrogen spectrum is an important piece of evidence for demonstrating an atom's quantized electronic structure.
2. When an electron jumps from a higher energy state to a lower energy state, an electromagnetic emission is obtained.

Step 3: Calculate the electron energy state

The wave number $\left(\overline{\nu }\right)$ of a spectral line in the emission spectrum of hydrogen, where,

$\left(\overline{\nu }\right)=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$, where, $n_1$= Initial energy state,$n_2$= Final energy state, and $R_H$= Rydberg's constant

Therefore, from the given information,

$\left(\overline{\nu }\right)=\frac{8}{9}{R}_H$

Therefore, from the above equation,

$\frac{8}{9}{R}_H=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$

$\Rightarrow \frac{8}{9}=\left(\frac{1}{1^2}-\frac{1}{{n_2}^2}\right)$

$n_2=3$

Hence, the correct option is (A).