The question is :-

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula $E = -2.178 \times 10^{-18} \frac{Z^2}{n^2}$

So $$E = -2.178 \times 10^{-18} .\frac{3^2}{1}$$ $$ E =-1.96 \times 10^{-17}~\mathrm{J}$$

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$.

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula

\begin{align} E &= -2.178 \times 10^{-18} \cdot \frac{Z^2}{n^2}.\\ E &= -2.178 \times 10^{-18} \cdot \frac{3^2}{1}\\ E &= -1.96 \times 10^{-17}~\mathrm{J} \end{align}

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$.

The question is :-

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula $E = -2.178 \times 10^{-18} \frac{Z^2}{n^2}$

So $$E = -2.178 \times 10^{-18} .\frac{3^2}{1}$$ $$ E =-1.96 \times 10^{-17}~\mathrm{J}$$

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$. Thanks in advance.

The question is :-

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula $E = -2.178 \times 10^{-18} \frac{Z^2}{n^2}$

So $$E = -2.178 \times 10^{-18} .\frac{3^2}{1}$$ $$ E =-1.96 \times 10^{-17}~\mathrm{J}$$

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$.

The question is :-

The ionization energy of $He^+$ is $19.6 \times 10^{-18}J$ $atom^{-1}$. What is the energy of the first stationary state (n=1) of $Li^{2+}$?

Since the question specifically states that n=1, I used the forumula $E = -2.178 \times 10^{-18} \frac{Z^2}{n^2}$

So $$E = -2.178 \times 10^{-18} .\frac{3^2}{1}$$ $$ E =-1.96 \times 10^{-17}J$$

But the correct answer is supposed to be $-4.14 \times 10^{-17}J$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set n at $ \frac{9}{4}$. Thanks in advance.

The question is :-

The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula $E = -2.178 \times 10^{-18} \frac{Z^2}{n^2}$

So $$E = -2.178 \times 10^{-18} .\frac{3^2}{1}$$ $$ E =-1.96 \times 10^{-17}~\mathrm{J}$$

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$. Thanks in advance.