Question

What transition of ${Li}^{+2}$ spectrum will have same wavelength as that of second line of Balmer series in ${He}^{+}$ spectrum?

Answer 1

As we know that,

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

For ${He}^{+}$-

$\frac{1}{\lambda }=R_H{\left(2\right)}^2(\frac{1}{2^2}-\frac{1}{4^2})$

For ${Li}^{+2}$-

$\frac{1}{\lambda }=R_H{\left(3\right)}^2(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$

Given that $\lambda$ is same.

Therefore,

$R_H{\left(3\right)}^2(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})=R_H{\left(2\right)}^2(\frac{1}{2^2}-\frac{1}{4^2})$

$\Rightarrow (\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})=(\frac{1}{3^2}-\frac{1}{6^2})$

Therefore,

$n_1=3$

$n_2=6$

Hence the transition is from $n=6$ to $n=3$.