Question

Find the second excitation energy of $L{i}^{2+}$ .
(A) 108.8 $eV$
(B) 81.6 $eV$
(C) 13.6 $eV$
(D) 95.2 $eV$

Answer 1

Hint: To solve this question one needs to know the atomic number of different elements. Also, by second excitation we mean that the electron is jumping from $1^{st}$ to $3^{rd}$ energy level. To find the second excitation energy of an element we subtract the energy of $1^{st}$ energy level from the energy of $3^{rd}$ energy level.

Complete Step by Step Solution:
It is given that an electron jumps from $1^{st}$ energy level to the $3^{rd}$ energy level i.e., second excited state.


We know that energy of $n^{th}$ energy level of an element is $-{\displaystyle \frac{13.6\times Z^2}{n^2}}$ where Z is the atomic number of the element and n is the energy level of the electron. This energy is in electron volts $(eV)$ . Atomic number is the number of protons in the nucleus of an atom.

Now to find the second excitation energy $\left(E\right)$ of $L{i}^{2+}$ we subtract the energy of $1^{st}$ energy level $$\left(E_1\right)$$ from the energy of $3^{rd}$ energy level $\left(E_3\right)$ .
Therefore, $E=E_3-E_1$
 $$E=-{\displaystyle \frac{13.6\times Z^2}{{n_3}^2}}-\left(-{\displaystyle \frac{13.6\times Z^2}{{n_1}^2}}\right)$$ ...(1)
Now we know that the atomic number of $Li$ is 3, so here we get $Z=3$ .
Also, $n_{{}_3}$ is the $3^{rd}$ energy level and $n_1$ is the $1^{st}$ energy level, so we get that $n_{{}_3}=3$ and $n_1=1$ .
Thus, equation (1) becomes,
$$E=-{\displaystyle \frac{13.6\times 3^2}{3^2}}-\left(-{\displaystyle \frac{13.6\times 3^2}{1^2}}\right)$$
$$E=-{\displaystyle \frac{13.6\times 3^2}{3^2}}+{\displaystyle \frac{13.6\times 3^2}{1^2}}$$
Taking $13.6\times 3^2$ common from both the terms we get,
 $E=-13.6\times 3^2\times \left({\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{1^2}}\right)$
 $E=-13.6\times 9\times \left({\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{1}}\right)$

Solving the brackets,
 $E=-13.6\times 9\times \left({\displaystyle \frac{1-9}{9}}\right)$
 $E=-13.6\times 9\times \left({\displaystyle \frac{-8}{9}}\right)$

Multiplying all the terms, we get
 $E=-13.6\times \left(-8\right)$
Thus, $E=108.8eV$
Hence, the correct option is A.

Note: First thing to keep in mind in all such questions is that second excited state is not the $2^{nd}$ energy level of the element, it is the $3^{rd}$ energy level. Thus, we can say that the $n^{th}$ excited state is $(n+1{)}^{th}$ energy level of any element. Also, the energy obtained by this method is in electron volts $(eV)$ and not in Joule $\left(J\right)$ . To convert $eV$ into $J$ use the conversion equation: $1eV=1.602\times {10}^{-19}J$ .