Question

What would be the wavelength and name of series respectively for the emission transition for $H$-atom if it starts from the orbit having radius $1.3225\,nm$ and ends at $211.6\,pm$?

• A. $434\,nm$, Balmer
• B. $434\,pm$, Paschen
• C. $545\,pm$, Pfund
• D. $600\,nm$, Lyman

Answer 1

Answer: (A)

Radius of $n^{th}$ orbit
$=\frac{0.529 \times n^{2} }{Z} \mathring{A}=\frac{52.9 \times n^{2}}{Z}pm$
Radius $\left(r_{2}\right)=1.3225\,nm=1322.5\,pm=\frac{52.9n^{2}_{2}}{Z}$
Radius $\left(r_{1}\right)=211.6\,pm=\frac{52.9\,n^{2}_{1}}{Z}$
$\therefore \frac{r_{2}}{r_{1}}=\frac{1322.5}{211.6}=\frac{n^{2}_{2}}{n^{2}_{1}}$
$\Rightarrow 6.25=\frac{n^{2}_{2}}{n^{2}_{1}}=\left(\frac{n_{2}}{n_{1}}\right)^{^2}$
$\Rightarrow \frac{n_{2}}{n_{1}}=\sqrt{6.25}$
$=2.5$
$\therefore n_{1}=2$, $n_{2}=5$ thus, the transition is from $5^{th}$ orbit to $2^{nd}$ orbit. It belongs to Balmer series.
$\bar{v}=R\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right)$
$\Rightarrow \bar{v} =1.097\times10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)$
$=1.097\times10^{7}\left(\frac{1}{4}-\frac{1}{25}\right)$
$=1.097\times10^{7}\times\frac{21}{100}$
$\lambda=\frac{1}{\bar{v}}=\frac{100}{1.097 \times 21 \times 10^{7}}m$
$=4.34\times10^{-7}\,m=434\times10^{-9}\,m$
$\lambda=434\,nm$
Thus, it lies in the visible region.