Question

Calculate the energy change required for an electron to move between states: a quantum jump up or down an energy-level diagram.

Short answer

The energy change required for an electron to jump from one state to another is$∆E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)Z_{eff}^{2}eV$

Step-by-step solution

Understanding the electron shells and energy levels.

Electrons of an atom always revolve following certain paths around the nucleus called orbits or call these Electron Shells. Each orbit has its own energy-specific energy levels. The orbit closer to the nucleus has a low energy level, and the orbit far away from the nucleus has high energy levels.

Understanding the relation between Energy level and principal Quantum number.

The principal Quantum number (n) can be represented as the specific orbit number belonging to a particular electron state.

There is a relationship between the energy level (E) of the electron and the closeness of the orbit to the nucleus that is given by,

$E=-\frac{13.6}{n^2}{Z}_{eff}^{2}eV$ ….. (1)

Here, is the atomic number.

Observe here that the energy level is negative with increasing n , which should always be a positive integer, i.e. $(n=1,2,3,\dots ,n)$. Hence the values of E will become as,

$\begin{array}{l}{E}_1=-13.6eV\\ E_2=-3.39eV\\ E_3=-1.51eV......\end{array}$

Hence, the energy level is quantized.

Understanding the change in Energy change required for an electron:

Consider a Hydrogen atom that has an atomic number (Z= 1) . The electron present in orbit with the quantum number $\left(n_1\right)$, which has the energy level $\left(E_1\right)$, gains energy to move to the next orbit with the quantum number $\left(n_2\right)$, which has the energy level $\left(E_2\right)$.

Then, the change in energy can be written from the equation (1) as:

$$\begin{gathered} ∆E=E_2-E_1 \\ =\frac{13.6}{n_2^2}{Z}_{eff}^{2}eV-\left(-\frac{13.6}{n_1^2}{Z}_{eff}^{2}eV\right) \\ =13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)Z_{eff}^{2}eV \end{gathered}$$

For elements other than Hydrogen, we can write equation (1) as,

$∆E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)Z_{eff}^{2}eV$