Question

How much energy is required to excite an electron in the ground state in a hydrogen atom to the third excited state? How much energy would it take to ionize the atom?

Answer 1

An electron in Bohr’s model of the atom is most stable at the ground state and most unstable at infinity. The energy in Bohr’s model is directly proportional to the square of atomic number and inversely proportional to the square of the number of orbits. We can substitute corresponding values to calculate energy at each level. The energy required is the change in energy.
Formulas used:
$E=\dfrac{-13.6{{z}^{2}}}{{{n}^{2}}}$
$\Delta E={{E}_{2}}-{{E}_{1}}$

Complete step-by-step solution:
In Bohr’s model of an atom, only select orbits or energy states are allowed in which the angular momentum of the electron is an integral multiple of $\dfrac{h}{2\pi }$
$L=n\dfrac{h}{2\pi }$
Here, $L$ is the angular momentum of the electron
$h$ is Planck’s constant
$n$ is the number of orbits
The energy of the electron in an orbit in Bohr’s model of an atom is given by-
$E=\dfrac{-13.6{{z}^{2}}}{{{n}^{2}}}$
Here, $E$ is the energy of the electron
$z$ is the atomic number
$n$ is the number of orbits
Given, initially, in a hydrogen atom, the electron is in ground state $\therefore n=1$.
In the above equation, we substitute given values to get,,
$\begin{aligned} & E=\dfrac{-13.6{{z}^{2}}}{{{n}^{2}}} \\\ & \Rightarrow {{E}_{1}}=\dfrac{-13.6\times {{1}^{2}}}{{{1}^{2}}} \\\ & \Rightarrow {{E}_{1}}=-13.6eV \\\ \end{aligned}$
Therefore, the energy of the electron in ground state is $-13.6eV$.
The energy of electron in third excited state or in $n=4$ is,
$\begin{aligned} & E=\dfrac{-13.6{{z}^{2}}}{{{n}^{2}}} \\\ & \Rightarrow {{E}_{2}}=\dfrac{-13.6\times {{1}^{2}}}{{{4}^{2}}} \\\ & \therefore {{E}_{2}}=-0.85eV \\\ \end{aligned}$
Therefore, the energy of electron in third excited state is $-0.85eV$
The energy required to move electron from ground state to third excited state=
$\begin{aligned} & \Delta E={{E}_{2}}-{{E}_{1}} \\\ & \Rightarrow \Delta E=-0.85-(-13.6)eV \\\ & \therefore \Delta E=12.75eV \\\ \end{aligned}$
Therefore, the energy required to move electrons from ground state to third excited state is $12.75eV$.
To ionize the atom means to completely remove the electron from the atom, i.e. move the electron to infinity. Therefore, $n=\infty$.
The energy of electron at $n=\infty$ is-
$\begin{aligned} & E'=\dfrac{-13.6}{{{\infty }^{2}}} \\\ & \Rightarrow E'=0eV \\\ \end{aligned}$
The energy required to completely remove the electron is,
$\begin{aligned} & \Delta E=E'-{{E}_{1}} \\\ & \Rightarrow \Delta E=0-(-13.6)eV \\\ & \Delta E=13.6eV \\\ \end{aligned}$
The energy required to ionize the atom is $13.6eV$.
Therefore, the energy required to excite the atom to the third excited state is $12.75eV$, and the energy required to ionize the atom is $13.6eV$.

Note: The total energy of the electron is negative due to very strong electrostatic and atomic forces acting on it. At infinity, there is no force acting on the electron and hence its energy is zero. When a change in energy is positive, it means that energy is given to the electron. When a change in energy is negative, it means that the energy is released by the electron.