Question

The first excitation energy of an electron of a hydrogen like atom is $40.8eV$. Find the energy needed to remove the electron to form the ion.

Answer 1

Hint: In order to solve this question, first we need to find the atomic number$Z$by equating the given $1st$excitation energy with the help of the formula mentioned below. Then put this value in the formula to get the answer.

Formula used:
${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV$\[
    \\
    \\
 \]where $Z = $atomic number of the atom, $n = $number of levels,
Excitation energy, $\Delta E = {E_n} - {E_{n - 1}}$

Complete step-by-step solution:
For a hydrogen like atom the the energy is given by
\[{E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV - - (i)\]
For energy level \[n = 1\]the energy is equal to:
\[{E_1} = \dfrac{{ - 13.6{Z^2}}}{{{{(1)}^2}}}eV - - (ii)\]
For energy level $n = 2$the energy is equal to:
${E_2} = \dfrac{{ - 13.6{Z^2}}}{{{{\left( 2 \right)}^2}}} - - (iii)$
Subtracting equation$(i)$from $(ii)$we get:
$
  \Delta E = {E_2} - {E_1} \\
  \Delta E = - 13.6{Z^2}\left( {\dfrac{1}{{{{\left( 2 \right)}^2}}} - \dfrac{1}{{{{\left( 1 \right)}^2}}}} \right) - - (iv) \\
 $
It is provided that $\Delta E = 40.8eV$. Now putting it in equation$(iv)$ we get:
$40.8 = - 13.6{Z^2}\left( {\dfrac{1}{{{{\left( 2 \right)}^2}}} - \dfrac{1}{{{{\left( 1 \right)}^2}}}} \right)$
$\therefore Z = 2$
To find the energy needed to remove the electron from ion we have:
$Z = 2$,$n = 2$
Then we put it in equation$(i)$we get:

$
  {E_2} = \dfrac{{ - 13.6{{\left( 2 \right)}^2}}}{{{{\left( 2 \right)}^2}}}eV \\
  {E_2} = - 54.4eV \\
 $
Energy needed to remove the electron to form the ion will be:
 $
   - \left( { - 54.4} \right)eV \\
  54.4eV \\
 $
The energy required to remove the electron to form the ion will be: $54.4eV$

Note:Some points to keep in mind while solving these types of problems:
Remember all the formulae of energies so that you should apply those in the right place.
You have to keep a negative sign in the final answer so that we get the energy needed to remove the electron to form the ion.