Question

An electron of a hydrogen like atom, having \(Z=4\), jumps from \(4^{\text {th }}\) energy state to \(2^{\text {nd }}\) energy state. The energy released in this process, will be : 
\((\) Given Rch \(=136 eV )\)
Where \(R =\) Rydberg constant 
\(c =\) Speed of light in vacuum 
\(h =\) Planck's constant

• A. $40.8 \,eV$
• B. $3.4 veV$
• C. $10.5 \,eV$
• D. $13.6 \, eV$

Answer 1

The Correct Option is A

Step 1: Recall the Formula for Energy Levels in Hydrogen-like Atoms
The energy of an electron in a hydrogen-like atom is given by: 

\( E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV} \)

where \( Z \) is the atomic number and \( n \) is the principal quantum number.

Step 2: Calculate the Energy Difference
The energy released (\( \Delta E \)) when an electron jumps from an initial state (\( n_i \)) to a final state (\( n_f \)) is given by the difference in energy levels:

\( \Delta E = E_{n_i} - E_{n_f} = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, \text{eV} \)

In this case, \( Z = 4 \), \( n_i = 4 \), and \( n_f = 2 \). Substituting these values, we get:

\( \Delta E = 13.6 \times (4^2) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \, \text{eV} \)

\( \Delta E = 13.6 \times 16 \left( \frac{1}{4} - \frac{1}{16} \right) \, \text{eV} \)

\( \Delta E = 13.6 \times 16 \left( \frac{4 - 1}{16} \right) \, \text{eV} \)

\( \Delta E = 13.6 \times 16 \times \frac{3}{16} \, \text{eV} \)

\( \Delta E = 13.6 \times 3 = 40.8 \, \text{eV} \)

Conclusion: The energy released in the process is 40.8 eV (Option 1).