Question

An electron in ${Li}^{2+}$ ion is in excited state $\left(n_2\right)$. The wavelength corresponding to a transition to second orbit is $48.24nm$. From the same orbit, wavelength corresponding to a transition to third orbit is $142.46nm$. The value of $n_2$ is :

• A. $5$
• B. $6$
• C. $7$
• D. $4$

Answer 1

Answer: (C)

$7$

$\frac{1}{\lambda }=R{z}^2$$(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

=$R{z}^2$ $(\frac{1}{2^2}-\frac{1}{n^2})$

${\frac{1}{\lambda }}_2=R{z}^2$ $(\frac{1}{3^2}-\frac{1}{n^2})$

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow$ $\frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{1}{4}-\frac{1}{n^2}}{\frac{1}{9}-\frac{1}{n^2}}$

$\frac{142.46}{4824}$=$\frac{\frac{n^2-4}{{4n}^2}}{\frac{n^2-9}{{9n}^2}}$

$3=\frac{9}{4}$ $\left(\frac{n^2-4}{n^2-9}\right)$

$4n^2-36=3n^2+2$

$n^2=48$

$n=\sqrt{48}$

$n\cong 7$

$\Rightarrow$value of $n_{2}is7$