Question

What is the potential energy of an electron present in the N shell of the $B{e}^{3+}$ ion?
A. -3.4 eV
B. -6.8 eV
C. -13.6 eV
D. -27.2 eV

Answer 1

Hint: To solve this question we have to remember the formula of total energy for a hydrogen like species in nth shell because here $B{e}^{3+}$ will behave as a hydrogen like species and the formula is $E_n=-13.6\left({\displaystyle \frac{Z^2}{n^2}}\right)eV/atom$. As we know that to find the potential in respect to energy we have a formula that is:
Potential energy = 2 $\times$ total energy

Complete step by step answer:
From your chemistry lessons you have learned about the total energy of an electron in the nth shell and also about the potential energy of an electron.
In order to solve this question firstly we have to know something about the ionization energy or ionization potential of an electron. Ionization energy is the amount of energy required to remove an electron from the valence or outermost shell of an isolated gaseous atom. And this process is an endothermic process. So, the formula to calculate the value is given as:
$$\mathrm{\Delta }E=13.6\left({\displaystyle \frac{Z^2}{n^2}}\right)eV$$

Where, Z = atomic number of the atom
n = Principle quantum number for the valence shell
Now, the total energy for an electron in the nth orbit is equal to its ionization energy for that shell but the difference is that the total energy is with negative sign and it is given as:
$$E_n=-13.6\left({\displaystyle \frac{Z^2}{n^2}}\right)eV$$

As we know that the total energy is the sum of kinetic energy and the potential energy of the electron in the nth orbit. And the formula for kinetic energy in the nth orbit is equals to:
$$K.E_n=-13.6\left({\displaystyle \frac{Z^2}{n^2}}\right)eV$$

And the potential energy of an electron is twice the kinetic energy. Therefore it is given as:
$$P.E_n=2\times -13.6\left({\displaystyle \frac{Z^2}{n^2}}\right)eV$$

Now, in the question we have to find the potential energy of the N shell of the and the principal quantum number for N shell is 4 (n = 4) and the atomic number of Be is 4 (Z = 4). So, after putting all the value in the formula of potential we will get:

$$P.E=2\times -13.6\left({\displaystyle \frac{4^2}{4^2}}\right)eV$$
$$\therefore P.E=-27.2eV/ion$$
So, the correct answer is “Option D”.

Note: If you will add the formulas of kinetic energy and the potential energy you will get the formula of total energy. The unit of energy is electron volt (eV) and here we have to find the potential energy of $B{e}^{3+}$ ion therefore the unit will be eV/ion. There are four shells that are K, L, M and N and the value of the principal quantum number for each shell is 1, 2, 3 and 4.