Question

The spectral line observed at $434\, nm$ 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 $n$ ?
$\left[\right.$ Rydberg constant, $\left.\left(R_{ H }\right)=109677 \,cm ^{-1}\right]$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

Given,

Rydberg constant $\left(R_{ H }\right)=109677 \,cm ^{-1}$

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

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(109677 \,cm ^{-1}\right)( 1 )^{2}\left(\frac{1}{(2)^{2}}-\frac{1}{\left(n_{2}\right)^{2}}\right) $

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

$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 $

$\Rightarrow n_{2}=5$