Question

The ground state energy of a hydrogen atom is $- 13.6eV$ . The energy of second excited state of $H{e^ + }$ ion in $eV$ is:
(A) $- 27.2$
(B) $- 3.4$
(C) $- 54.4$
(D) $- 6.04$

Answer 1

In this question it is given that the ground state energy of hydrogen atom as $- 13.6eV$ but we have to find the energy in the second excited state of $H{e^ + }$ ion with the help of the formula for energy. We must know the atomic number of helium atoms and use this in the formula for energy.
Use formula: $E = {E_0}\dfrac{{{Z^2}}}{{{n^2}}}eV$

Complete answer:
According to the given information we have ground state energy of hydrogen atom (as ${E_0}$ )
Therefore, ${E_0}$ is $- 13.6eV$
Energy levels are numbered as for ground state $n = 1$ , first excited state is $n = 2$ , second excited state is $n = 3$ , and so on.
Now, let us use the formula for energy calculation in the energy levels of the $H{e^ + }$ ion as given below:
$E = {E_0}\dfrac{{{Z^2}}}{{{n^2}}}eV$
Here, for helium ion $Z = 2$ and for second excited state $n = 3$ and put these values in above equation, we get
$E = - 13.6 \times \dfrac{{{{(2)}^2}}}{{{{(3)}^2}}}eV$
$\Rightarrow E = - 13.6 \times \dfrac{4}{9}eV$
$\Rightarrow E = - 6.04eV$
The energy at the second excited state of $H{e^ + }$ ion is $- 6.04eV$
Thus, the correct answer is option D.

Note:
In these kinds of questions, if the ground state energy is given then we know that it is enough to find out the energy of any state of any ion at any energy level. Along with the ground state energy of the hydrogen atom, the energy of which ion we want to find we must know the atomic number as well as which excited state’s energy we want to find the energy.