Question

The spectral line observed at $434nm$ in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the $n$ th orbit. What is the value of $n$ ?
$[$ Rydberg constant, $\left(R_H\right)=109677c{m}^{-1}]$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

Given,

Rydberg constant $\left(R_H\right)=109677c{m}^{-1}$

Wavelength $(\lambda )$ of the line in the Balmer series is $=434nm$

Atomic number $(Z)$ of hydrogen is $=1$

Integer value of $n_1=2$

From Rydberg equation,

$\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\frac{1}{434\times 10^{-7}cm}=\left(109677c{m}^{-1}\right)(1{)}^2\left(\frac{1}{(2{)}^2}-\frac{1}{{\left(n_2\right)}^2}\right)$

$\frac{1}{434\times 10^{-7}cm}=\left(109677c{m}^{-1}\right)(1{)}^2\left(\frac{1}{4}-\frac{1}{n_2^2}\right)$

$23041.47=109677\times \frac{1}{4}-109677\times \frac{1}{n_2^2}$

$-4377.78=-109677\times \frac{1}{n_2^2}$

$n_2^2=\frac{1}{0.0399}$

$\Rightarrow n_2^2=25.06$

$\Rightarrow n_2=5$