© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Formula
The energy of an electron in the n-th level of a hydrogen-like atom is given by:
$$
E_n = -\frac{13.6\,\text{eV} \times Z^2}{n^2}
$$
where:
$13.6\,\text{eV}$ is the binding energy of the electron in the ground state of a hydrogen atom.
$Z$ is the atomic number of the ion (for $Li^{++}$, $Z = 3$).
$n$ is the principal quantum number corresponding to the energy level.
Step 2: Determine the Principal Quantum Number (n)
The question specifies the first excited state. For hydrogen-like atoms:
Ground state corresponds to $n = 1$.
First excited state corresponds to $n = 2$.
Thus, $n = 2$.
Step 3: Substitute the Known Values
Substitute $Z = 3$ and $n = 2$ into the formula:
$$
E_n = -\frac{13.6 \,\text{eV} \times (3)^2}{(2)^2}
$$
Step 4: Calculate the Numerical Value
Simplify inside the formula step-by-step:
$Z^2 = (3)^2 = 9$
$n^2 = (2)^2 = 4$
$$
E_n = -\frac{13.6 \,\text{eV} \times 9}{4}
= -\frac{122.4\,\text{eV}}{4}
= -30.6\,\text{eV}
$$
The negative sign here indicates that the electron is bound to the nucleus.
Step 5: Interpret the Result
The magnitude of this energy ($30.6\,\text{eV}$) is precisely the energy required to remove the electron from the first excited state of $Li^{++}$. Therefore, the energy needed is:
$30.6\,\text{eV}$.
