Question

The wavelength (in $\mathring{A}$ ) of an emission line obtained for $L{i}^{2+}$ during a electronic transition from $n_2=2$ to $n_1=1$ is: $(R=$ Rydberg constant $)$

• A. $3R/4$
• B. $27R/4$
• C. $4/3R$
• D. $4/27R$

Answer 1

Answer: (D)

$\frac{1}{\lambda }=R_H(Z{)}^2\left(\frac{1}{{\left(n_1\right)}^2}-\frac{1}{{\left(n_2\right)}^2}\right)$

where, $R_H=R,Z=3($ for $L{i}^{2+},$ atomic no. $3)$

$n_2=2,n_1=1$

$\frac{1}{\lambda }=R(3{)}^2\left(\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}\right)$

$=\frac{R\times 9\times 3}{4}$

$\Rightarrow \lambda =\frac{4}{27R}$