Question

The ratio of the wavelengths for $2\to 1$ transition in $L{i}^{2+},{\text{He}}^{+}$ and $H$ is

• A. $1:2:3$
• B. $\frac{1}{9}:\frac{1}{4}:\frac{1}{1}$
• C. $1:4:1$
• D. $3:2:1$

Answer 1

Answer: (B)

From Bohr’s formula , the wave number $\left(\frac{1}{\lambda }\right)$ is given by
$\frac{1}{\lambda }=Z^{2}R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where Z is atomic number, R the Rydberg's constant and n is the quantum number.
$\Rightarrow \lambda \propto \frac{1}{{\text{Z}}^2}$
Atomic number of lithium is 3, of helium is 2 and of hydrogen is 1.
$\therefore {\left(\lambda \right)}_{{\left(\text{Li}\right)}^{2+}}:{\left(\lambda \right)}_{{\left(\text{He}\right)}^{+}}:{\left(\lambda \right)}_{\text{H}}=\frac{1}{{\left(3\right)}^2}:\frac{1}{{\left(2\right)}^2}\text{:}1$
$=\frac{1}{9}\text{:}\frac{1}{4}\text{:}1$