Question

The possible energy values of the excited states for electrons in bohr’s first orbit of hydrogen is:
A.$- 3.4eV$
B.$- 4.2eV$
C.$- 6.8eV$
D.$+ 6.8eV$

Answer 1

Excited state is defined as the state with energy higher than the ground state or we can say the quantum state of a system that has a higher energy than the ground state is called an excited state. The concept was explained by Bohr.

Complete step-by-step answer:
Bohr was the scientist who give the systematic description of an atom especially the hydrogen. He was the first one to incorporate the theory of quantum or we can say the quantum theory. He stated that the atoms move around the nucleus in well defined orbits. These orbits are further called as the quantum numbers. We can further say that the electrons jump to higher or lower levels. Now you must be thinking how then we must know that it happens due to the absorption and emission of energy. Now in the question we are discussing about the excited state now we will define what excited state is.
Excited state is defined as any state or we can say the quantum state of a system that has a higher energy than the ground state. Further ground state is the state where we can say the hydrogen atom has the single electron in its lower energy state.
$E = - 13.6 \times \dfrac{{{Z^2}}}{{{n^2}}}$
Where $\dfrac{{{Z^2}}}{{{n^2}}}$ is the ratio of atomic number to the quantum number.
Then we will calculate the energy value for the first excited state
${E_2} = \dfrac{{ - 13.6}}{4} = - 3.4eV$
For the next excited state
${E_2} = \dfrac{{ - 13.6}}{9} = - 1.5eV$

Hence, the correct answer for the given question is option (A).

Note: The calculation of every excited state proper calculation should be carried out. The energy for the second excited state is $- 1.5eV$, third excited state energy is $- 0.85eV$ and the decreasing trend in value is observed for all excited states where n is a natural number (value of principal quantum number).