Chemistry - Atomic Structure Question with Solution | TestHub

Using Rydberg's equation for the transition of the electron,

$\frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}} \left({\mathrm{Z}}^2\right) \left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

Where, 
${\mathrm{R}}_{\mathrm{H}}=1.097\times {10}^7 {\mathrm{m}}^{-1}$

$\mathrm{Z}=6$ for carbon

${\mathrm{n}}_1=1, {\mathrm{n}}_2=3$

$\frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\times (6{)}^2\left[\frac{1}{1^2}-\frac{1}{3^2}\right] {\mathrm{m}}^{-1}$

$\frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\times 36\left[\frac{9-1}{9}\right] {\mathrm{m}}^{-1}$

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\times 36\times \frac{8}{9} {\mathrm{m}}^{-1} \\ \frac{1}{\mathrm{\lambda }}=35.104\times {10}^7 {\mathrm{m}}^{-1} \end{gathered}$$

$$\begin{gathered} \mathrm{\lambda }=\frac{1}{35.104\times {10}^7} \mathrm{m} \\ \mathrm{\lambda }=0.0285\times {10}^{-7} \mathrm{m} \\ \mathrm{\lambda }=2.85\times {10}^{-9} \mathrm{m} \\ \mathrm{\lambda }=2.85 \mathrm{nm} \end{gathered}$$