Question

In a hydrogen atom, if energy of an electron in the ground state is $13.6eV$ , then that in the $2nd$ excited state is:
$(A)1.51ev$
$(B)3.4ev$
$(C)6.04ev$
$(D)13.6ev$

Answer 1

We know that, according to Bohr’s atomic model in a hydrogen atom, the energy of each electron in each shell is fixed. In the solution, we will see the formula to calculate energy in different states/shells in the atom. Excited state is the state in which an electron jumps when it gains some energy. Then, the electron comes back to its initial state or shell by losing energy in the form of color.

Complete answer:
We know that according to Bohr’s atomic model in a hydrogen atom, the energy of each electron in each shell is fixed.
The formula for the energy of an electron in the orbital is given as:
${E_n} = \dfrac{{13.6 \times {Z^2}}}{{{n^2}}}$
Where, Z is the atomic number of the element and n is the number of shells in which the electron is present.
For, hydrogen atom the value of Z is $1.$ Second excited level here refers to $n = 3.$
Now, put these values in the formula mentioned above:
${E_n} = \dfrac{{13.6 \times {1^2}}}{{{3^2}}}$
${E_n} = 1.51ev$
Therefore, the correct option is $(A)1.51ev$ .

Note:
Bohr’s atomic model represents the structure of an atom as there is a small nucleus which is positively charged and the electrons (which are negatively charged) revolve around it. It states that electrons move in orbitals with fixed energy levels. It explains that electrons jump in the higher energy level after gaining energy and come back to their ground state by losing the energy.