Question

Compare the energy released during direct electronic transition from 4^th orbit to 1^st orbit with the energy released during successive electronic transitions between the same orbits in H-atom. Is the energy of all successive transitions between the 4^th orbit and 1^st orbit equal?

Answer 1

1. When an electron moves from one orbit to another in a hydrogen atom, the energy released is calculated as follows:

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

2. The energy released during direct electronic transition from 4^th orbit to 1^st orbit;

$$\begin{gathered} ∆E=-13.6\left(\frac{1}{1_2^2}-\frac{1}{4_1^2}\right)eV \\ ∆E=12.75eV \end{gathered}$$

3. Now, the energy is released when the successive electronic transitions between the fourth orbit to the first orbit occur.

( successive transition means electron jumps from $4\to 3\to 2\to 1$)

4. Consider; $E_1,E_2,E_3$ are the energies released when the electron jumps from $4\to 3\to 2\to 1$respectively.

$$\begin{gathered} E_1=-13.6\left(\frac{1}{3_{}^2}-\frac{1}{4_{}^2}\right)eV \\ {E}_1=-0.66eV \end{gathered}$$

$$\begin{gathered} E_2=-13.6\left(\frac{1}{2_{}^2}-\frac{1}{3_{}^2}\right)eV \\ {E}_2=-1.89eV \end{gathered}$$

$$\begin{gathered} E_3=-13.6\left(\frac{1}{1_{}^2}-\frac{1}{2_{}^2}\right)eV \\ {E}_3=-10.2eV \end{gathered}$$

5. Now, $E_1+E_2+E_3=-0.66eV-1.89eV-10.2eV=-12.75eV$

6. This is the amount of energy released during the direct transition from the fourth to the first orbit.

7. However, the magnitude of energy released increases with each transition.