Question

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer 1

$r=0.529\times n^2{A}^o$

$r_2=1.3225nm\times 10A^o/nm=0.529\times n_2^2{A}^o$

$n_2^2=25$

$n_2=5$

$r=0.529\times n^2{A}^o$

$r_1=211.6pm\times 0.01A^o/pm=0.529\times n_1^2{A}^o$

$n_1^2=4$
$n_1=2$

The transition $n_2\to n_1$ is $5\to 2$. It is in the visible region of light. It belongs to Balmer series.

$\frac{1}{\lambda }=109677.58c{m}^{-1}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=109677.58c{m}^{-1}[\frac{1}{2^2}-\frac{1}{5^2}]$

$\frac{1}{\lambda }=23032c{m}^{-1}$

$\lambda =4.34\times {10}^{-5}cm$