Question

If $\lambda$ is the wavelength of hydrogen atom from the transition $n=3$ to $n=1$, then what is the wavelength for doubly ionised lithium ion for same transition?

• A. $\frac{\lambda }{3}$
• B. $3\lambda$
• C. $\frac{\lambda }{9}$
• D. $9\lambda$

Answer 1

The correct option is C $\frac{\lambda }{9}$

For wavelength:
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Here, transition is same.

So, $\lambda \propto \frac{1}{Z^2}$

$\frac{{\lambda }_H}{{\lambda }_{Li}}=\frac{(Z_{Li}{)}^2}{(Z_H{)}^2}=\frac{(3{)}^1}{(1{)}^2}=9$

${\lambda }_{Li}=\frac{{\lambda }_H}{9}=\frac{\lambda }{9}$