Question

If the binding energy of the electron in a hydrogen atom is $13.6eV$ then the energy required to remove the electron from the first excited state of $L{i^{ + + }}$ is
(A) $122.4eV$
(B) $30.6eV$
(C) $13.6eV$
(D) $3.4eV$

Answer 1

Hint
To solve this question, we have to use the formula for the total energy of an electron in the nth orbit. For removing the electron from an atom, we have to make its total energy zero.
Formula Used: The formula used in solving this question is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}}$
Here ${E_n}$ is the energy in electron-volts for an electron in the nth state of an atom having atomic number equal to $z$.

Complete step by step answer
We know that the energy of an electron in the nth state of an atom is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}}$ ………………….(1)
As the atom given in the question is Lithium, which has an atomic number of $3$, so we have $z = 3$.
Also, the ground state corresponds to $n = 1$. So the first excited state for an atom will correspond to the $n = 2$ state. So we have $n = 2$. Substituting these values in (1), we get
$\Rightarrow {E_n} = - 13.6\dfrac{{{3^2}}}{{{2^2}}}$
$\Rightarrow {E_n} = - 30.6eV$
Therefore the energy of the electron in the first excited state of $L{i^{ + + }}$ is equal to $- 30.6eV$.
For removing the electron, we have to make it free from the effect of the nucleus. That is, we need to make its total energy zero. For this, let the energy supplied to the electron be equal to $E$. So we have
$\Rightarrow {E_n} + E = 0$
$\Rightarrow - 30.6 + E = 0$
So we get the energy supplied as
$\Rightarrow E = 30.6eV$
Thus the energy which is required to remove the electron from the first excited state of $L{i^{ + + }}$ is equal to $30.6eV$.
Hence, the correct answer is option B.

Note
If we remember the formula for the total energy of an electron in the nth state, then we do not require the value of the binding energy of the electron in the hydrogen atom, which is given in the question. But if we do not remember it, then we can use this value, as it corresponds to the energy in the ground state of hydrogen $n = 1$, which has $z = 1$.