Thank you for reporting, we will resolve it shortly
Q.
Which transition of $Li ^{2+}$ is associated with same energy change as $n =6$ to $n =4$ transition in $He ^{+}$?
Structure of Atom
Solution:
In $He ^{\oplus} ; n =6$ the corresponding energy level in $Li ^{2+}$ ion will be
$\begin{bmatrix}\text { For } He ^{\oplus} \\ z=2\end{bmatrix} \rightarrow E _{6}$
$=\frac{-1312}{(6)^{2}} \times(2)^{2}$
$=-\frac{1312}{36} \times 4=\frac{-1312}{9}$
Corresponding energy level for $Li^{2+}_{(z=3)} = \frac{1312}{n^2} \times(3)^2$
$= - \frac{1312}{9} n^2 = 81$
$n =9$
In $He ^{\oplus} n =4$; the corresponding energy level in $Li ^{2+}$ ions
For $He ^{\oplus}_{( z =2)} E _{4}=-\frac{1312}{(4)^{2}} \times 4=-\frac{1312}{4}$
Corresponding energy level for $Li^{\oplus}_{(z=3)}=\frac{-1312}{n^{2}}\times(3)^2$
$= -\frac{1312}{4}$
$n^{2}=36\,\, n=6$
Shortcut
$\frac{1}{\lambda_{ He ^{2+}}}=R \times(2)^{2}\left[\frac{1}{4^{2}}-\frac{1}{6^{2}}\right]$ ...(1)
$\frac{1}{\lambda_{ L ^{2+}}}=R \times(3)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ ...(2)
For same energy equation (1) must be equal to equation (2) which only possible when $n _{1}=6$ and $n _{2}=9$