Question

Which transition of $L{i}^{2+}$ is associated with same energy change as $n=6$ to $n=4$ transition in $H{e}^{+}$?

• A. n = 3 to n = 1
• B. n = 8 to n = 6
• C. n = 9 to n = 6
• D. n = 2 to n = 1

Answer 1

Answer: (C)

In $H{e}^{\oplus };n=6$ the corresponding energy level in $L{i}^{2+}$ ion will be
$\left[\begin{array}{c}\text{ For }H{e}^{\oplus }\\ z=2\end{array}\right]\to E_6$
$=\frac{-1312}{(6{)}^2}\times (2{)}^2$
$=-\frac{1312}{36}\times 4=\frac{-1312}{9}$
Corresponding energy level for $L{i}_{(z=3)}^{2+}=\frac{1312}{n^2}\times (3{)}^2$
$=-\frac{1312}{9}{n}^2=81$
$n=9$
In $H{e}^{\oplus }n=4$; the corresponding energy level in $L{i}^{2+}$ ions
For $H{e}_{(z=2)}^{\oplus }{E}_4=-\frac{1312}{(4{)}^2}\times 4=-\frac{1312}{4}$
Corresponding energy level for $L{i}_{(z=3)}^{\oplus }=\frac{-1312}{n^2}\times (3{)}^2$
$=-\frac{1312}{4}$
$n^2=36n=6$
Shortcut
$\frac{1}{{\lambda }_{H{e}^{2+}}}=R\times (2{)}^2\left[\frac{1}{4^2}-\frac{1}{6^2}\right]$ ...(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 $n_2=9$