A hydrogen like atom number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition ot quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy of hydrogen atom is -13.6 ev.

Explanation

To solve the problem step by step, we will use the formulas for the energy levels of a hydrogen-like atom and the information provided in the question.

### Step 1: Understand the Energy Level Formula
For a hydrogen-like atom, the energy of an electron in a quantum state with principal quantum number \( n \) is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number.

### Step 2: Maximum Energy Photon Emission
The maximum energy photon emitted when transitioning from the excited state \( 2n \) to the ground state \( n = 1 \) is given as 204 eV. Therefore, we can write:
\[ 204 = \left| E_1 - E_{2n} \right| \]
Substituting the energy level formula:
\[ 204 = \left| -\frac{13.6 Z^2}{1^2} - \left(-\frac{13.6 Z^2}{(2n)^2}\right) \right| \]
This simplifies to:
\[ 204 = 13.6 Z^2 \left(1 - \frac{1}{4n^2}\right) \]

### Step 3: Energy for Transition to State n
The energy emitted when transitioning from \( 2n \) to \( n \) is given as 40.8 eV. Thus:
\[ 40.8 = \left| E_n - E_{2n} \right| \]
Substituting the energy level formula:
\[ 40.8 = \left| -\frac{13.6 Z^2}{n^2} - \left(-\frac{13.6 Z^2}{(2n)^2}\right) \right| \]
This simplifies to:
\[ 40.8 = 13.6 Z^2 \left(\frac{1}{4n^2} - \frac{1}{n^2}\right) \]
This can be rewritten as:
\[ 40.8 = 13.6 Z^2 \left(-\frac{3}{4n^2}\right) \]

### Step 4: Solve for Z and n
From the first equation:
\[ 204 = 13.6 Z^2 \left(1 - \frac{1}{4n^2}\right) \]
From the second equation:
\[ 40.8 = -\frac{40.8 \cdot 4n^2}{13.6 Z^2 \cdot 3} \]
Rearranging gives:
\[ Z^2 = \frac{40.8 \cdot 4n^2}{13.6 \cdot 3} \]

### Step 5: Substitute Z^2 into the First Equation
Substituting \( Z^2 \) from the second equation into the first:
\[ 204 = 13.6 \left(\frac{40.8 \cdot 4n^2}{13.6 \cdot 3}\right) \left(1 - \frac{1}{4n^2}\right) \]
This simplifies to:
\[ 204 = \frac{40.8 \cdot 4n^2}{3} \left(1 - \frac{1}{4n^2}\right) \]
Solving this equation will yield the values of \( n \) and \( Z \).

### Step 6: Calculate Ground State Energy
Once \( Z \) is found, the ground state energy can be calculated using:
\[ E_1 = -\frac{13.6 Z^2}{1^2} \]

### Step 7: Calculate Minimum Energy of Hydrogen Atom
The minimum energy of a hydrogen atom is given as \( -13.6 \) eV, which is the energy of the ground state.

### Final Values
After solving the equations, we find:
- \( n = 2 \)
- \( Z = 4 \)
- Ground state energy \( E_1 = -\frac{13.6 \cdot 16}{1} = -217.6 \) eV.