Question

Review with derivation of the Bohr model of the hydrogen atom and apply this reasoning to predict the energy levels of ionized helium ${\mathbf{He}}^{\mathbf{+}}$ (a helium atom with only one electron, and a nucleus containing two protons and two neutrons). What are the energies in of the ground state and the first excited state? What is the energy of a photon emitted in a transition from the first excited state to the ground state? How do these results differ from those for a hydrogen atom?

Short answer

• The energy of the electron in the ground state and first excited state is$-54.4\text{eV}$and$-13.6\text{eV}$.
• The energy of emitted photon is$40.8\text{eV}$.

Step-by-step solution

Definition of angular acceleration

The rate of change of the angular velocity with respect to time is known as the angular acceleration.

Find the energy of the electron in the first excited state

The expression for energy of electron form Bohr’s theory is given as follows:

$\mathrm{E}=\frac{\left(-13.6\right){\mathrm{Z}}^2\text{eV}}{{\mathrm{N}}^2}$

Here, N represents the orbit number and Z is the atomic number.

For ground state:

Substitute 1 for N and 2 for Z.

$\begin{array}{rcl}{\mathrm{E}}_1& =& \frac{\left(-13.6\right){\left(2\right)}^2\text{eV}}{{\left(1\right)}^2}\\ & =& -54.4\text{eV}\\ & & \end{array}$

Therefore, the ground state energy of the electron in Helium atom is$-54.4\text{eV}$.

For first excited state:

Substitute 2 for N and 2 for Z in the expression

$\begin{array}{rcl}{\mathrm{E}}_1& =& \frac{\left(-13.6\right){\mathrm{Z}}^2\text{eV}}{{\mathrm{N}}^2}\\ {\mathrm{E}}_2& =& \frac{\left(-13.6\right){\left(2\right)}^2\text{eV}}{{\left(2\right)}^2}\\ & =& -13.6\text{eV}\\ & & \end{array}$

Therefore, the energy of the electron in the first excited state is $-13.6\text{eV}$.

 Step 3: Find the energy of emitted by photon

When an electron makes transition from the first excited state to ground state, the energy of the emitted photon is given as follows:

$\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1$

Substitute

$\begin{array}{rcl}{\mathrm{E}}_1& =& -54.4\text{eV}\\ {\mathrm{E}}_2& =& -13.6\text{eV}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{E}& =& \left(-13.6\right)-\left(-54.4\mathrm{ev}\right)\\ & =& 40.8\text{eV}\\ & & \end{array}$

Hence, the energy of emitted photon is$40.8\text{eV}$.