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

To solve this problem, we need to determine the energy released when an electron in a hydrogen-like atom with atomic number \(Z = 4\) transitions from the \(4^{\text{th}}\) energy state to the \(2^{\text{nd}}\) energy state. The energy levels of a hydrogen-like atom are given by the formula:

\(E_n = -\frac{Z^2 \cdot Rch}{n^2}\), where \(n\) is the principal quantum number.

Given:

• \(Z = 4\)
• \(Rch = 136 \, \text{eV}\)

First, calculate the energy of the \(4^{\text{th}}\) energy level (\(E_4\)):

\(E_4 = -\frac{4^2 \cdot 136}{4^2} = -\frac{16 \cdot 136}{16} = -136 \, \text{eV}\)

Next, calculate the energy of the \(2^{\text{nd}}\) energy level (\(E_2\)):

\(E_2 = -\frac{4^2 \cdot 136}{2^2} = -\frac{16 \cdot 136}{4} = -544 \, \text{eV}\)

The energy released when the electron transitions from the \(4^{\text{th}}\) to the \(2^{\text{nd}}\) energy level is the difference in energy between these two levels:

\(\Delta E = E_2 - E_4 = (-544 \, \text{eV}) - (-136 \, \text{eV}) = -544 + 136 = -408 \, \text{eV}\)

The energy released is \(40.8\, \text{eV}\), as energy release is taken as a positive value:

Therefore, the correct answer is 40.8 eV.