Question
Question: Which of the following is the energy of a possible excited state of Hydrogen? A. +6.8 eV B. +13....
Which of the following is the energy of a possible excited state of Hydrogen?
A. +6.8 eV
B. +13.6 eV
C. -6.8 eV
D. -3.4 eV
Solution
Hint-We would be using the concepts of the Electronic Energy of molecules. As stated by Neil Bohr who was the first person to explain the distribution of electrons around the nucleus. The electronic energy of a Hydrogen atom has a negative sign that depicts that the energy of an electron in an atom is less than the energy of a free electron at any possible excited state.
Formula used:
Energy of an electron in the nth orbit is
E=−n213.5×z2eV
Complete step by step answer:
Step 1:
From the above formula the energy of a Hydrogen atom is determined by the formula
E=n2E0
Here,E0=−13.6 eV
Step 2:
As we know, that the excited state of a Hydrogen atom can only be negative then all the positive values should be eliminated. Thus, options A and D are eliminated from consideration.
Step 3:
Now, let’s find the energy of the different energy levels,
For n=1, E=12−13.6=−13.6eV
For n=2, E=22−13.6=4−13.6=−3.4eV
For n=3, E=32−13.6=9−13.6=−1.51eV
For n=4, E=42−13.6=16−13.6=−0.85eV
For n=5,E=52−13.6=2513.6=−0.55eV
Step 4
From all these values, the energies of an electron in a Hydrogen atom in the first, second, third, fourth, and fifth energy levels are−13.6 eV, −3.4 eV, −1.51 eV, −0.85 eV, and 0.55 eV.
So, from the above question, the answer is option D, the energy of a possible excited state of a Hydrogen atom is−3.4 eV.
Note:We must remember that the total excited state of a Hydrogen atom will always be negative. And the unit of the energies will be measured in electron Volt (eV). The excited state of the hydrogen molecule is when it absorbs a certain amount of energy which is equivalent to E1. So, the electron moves from the orbital of the first shell i.e. when n = 1 towards the orbital of the second shell i.e. n = 2.