Question

What is the potential energy of electrons in the L-shell of the hydrogen atom?
(a) -13.6 eV
(b) -6.8 eV
(c) -10.2 eV
(d) -1.7 eV

Answer 1

To solve this question we need to remember that the magnitude of the total energy of an electron in nth shell is inversely proportional to the square of the principal quantum number and directly proportional to the square of the number of protons present in the nucleus of the atom. The magnitude of the potential energy of an electron in nth orbit is twice that of the total energy.

Complete step by step answer:
For solving this question we need to first understand the ionization energy for an electron. It is also called the ionization potential. It refers to the amount of energy that needs to be supplied in order to remove an electron from the outermost shell/valence shell of an isolated gaseous atom to infinity. It is generally an endothermic process. Its value can be calculated by using the formula:
$\Delta E=13.6\times \cfrac { { Z }^{ 2 } }{ { n }^{ 2 } } eV$

Where ‘n’ is the principal quantum number for the valence shell and ‘Z’ is the number of protons present in the nucleus of the atom i.e. its atomic number. The total energy of an electron in nth orbit is equal to its ionization potential for that orbit but with a negative sign.
Hence the total energy of an electron in its nth orbit will be:
${ E }_{ n }=-13.6\times \cfrac { { Z }^{ 2 } }{ { n }^{ 2 } } eV$

This total energy is actually the sum of the kinetic energy of the electron in its nth orbit as well as the potential energy of the electron in its nth orbit. The kinetic energy of an electron in the nth orbit is equal to:
$K.{ E }_{ n }=-13.6\times \cfrac { { Z }^{ 2 } }{ { n }^{ 2 } } eV$
While the potential energy for an electron in its nth orbit is equal to:
$P.{ E }_{ n }=-2\times 13.6\times \cfrac { { Z }^{ 2 } }{ { n }^{ 2 } } eV$
Which gives us the total energy as ${ E }_{ n }=-13.6\times \cfrac { { Z }^{ 2 } }{ { n }^{ 2 } } eV$

For L-shell, the principal quantum number is 4 i.e. the value of n = 4. For hydrogen atoms, the value of Z is 1. Hence the potential energy of the electron in its fourth orbit will be:
$P.{ E }_{ 4 }=-2\times 13.6\times \cfrac { 1 }{ { 4 }^{ 2 } } eV=-1.7 eV$
So, the correct answer is “Option D”.

Note: Always remember that the energy of various subshells present within a shell in a Hydrogen atom are degenerate because hydrogen atom is a single electron system due to which there are no electron-electron repulsions present in the system.