Question

The energy difference between the first two levels of hydrogen atom is 10.2 eV. What is the corresponding energy difference for a singly ionized helium atom?
A. 10.2 eV
B. 81.6 eV
C. 204 eV
D. 40.8 eV

Answer 1

Hint:The singly ionized helium atom is having one electron in the orbit and the hydrogen atom is also having one electron in its orbit. So the singly ionized helium atom is hydrogen like atom.

Formula used:
\[\Delta E = \left( {13.6eV} \right){Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\]
Here Z is the atomic number of the hydrogen-like atoms.

Complete step by step solution:
The first two levels of a hydrogen atom are the ground state and the first excited state. The principal quantum number of the ground state is 1 and the principal quantum number of the first excited state is 2.
\[{n_1} = 1\]
\[\Rightarrow {n_2} = 2\]
The energy difference between two energy level of the hydrogen like atom is given by the Bohr’s model as,
\[\Delta E = \left( {13.6eV} \right){Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\]
Here Z is the atomic number of the hydrogen-like atoms.

The hydrogen-like atom is an atom which has a single electron in its orbit. The atomic number of helium atom is 2, i.e. there is 2 electrons in a neutral helium atom. When it is a singly ionized atom then it loses one electron and there is only one electron in the orbit of the helium atom and it is hydrogen like atom.

For hydrogen, the atomic number is 1. Using energy difference formula for hydrogen atom,
\[10.2eV = \left( {13.6eV} \right)\left( {{1^2}} \right)\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\]
\[\Rightarrow 10.2eV = \left( {13.6eV} \right)\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \ldots \left( 1 \right) \\ \]
Using energy difference formula for helium atom,
\[\Delta E = \left( {13.6eV} \right)\left( {{2^2}} \right)\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\ \]
\[\Rightarrow \Delta E = \left( {13.6eV} \right)4\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \ldots \left( 2 \right)\]

Dividing both the equations, we get
\[\dfrac{{10.2eV}}{{\Delta E}} = \dfrac{{\left( {13.6eV} \right)\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)}}{{\left( {13.6eV} \right)4\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)}} \\ \]
\[\Rightarrow \dfrac{{10.2eV}}{{\Delta E}} = \dfrac{1}{4} \\ \]
\[\Rightarrow \Delta E = 4 \times 10.2\,eV \\ \]
\[\therefore \Delta E = 40.8\,eV\]
Hence, the energy difference in the helium is 40.8 eV.

Therefore, the correct option is D.

Note: According to the Bohr’s model, the energy of each orbit is fixed. The electron has minimum energy in the ground state. When energy is given to the atom then the electron gains energy and jumps to a higher energy level. The maximum energy required to remove an electron from the ground state out of the atom is called the ionization energy.