A hydrogen - like atom of atomic number `Z` is in an excited state of quantion number `2 n` it can emit a maximum energy photon of energy `40.8 eV` is emitted find n , Z and the ground state energy (in eV) for this atom . Also calculate the minimum energy (in eV) that can be emitted by this atom during de- exclation , Ground state energy of hydrogen atom is - `13.6 eV`

Explanation

To solve the problem step by step, we will first analyze the energy levels of a hydrogen-like atom and then apply the given information to find the values of \( n \), \( Z \), the ground state energy, and the minimum energy that can be emitted during de-excitation.

### Step 1: Understanding the Energy Levels
For a hydrogen-like atom, the energy of the electron in the \( n \)-th energy level is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( Z \) is the atomic number.

### Step 2: Maximum Energy Photon Emission
The maximum energy photon emitted when transitioning from an excited state \( 2n \) to the ground state \( n=1 \) is given as \( 40.8 \, \text{eV} \). This can be expressed as:
\[ E_{\text{max}} = E_1 - E_{2n} \]
Substituting the energy levels:
\[ 40.8 = -\frac{13.6 \, Z^2}{1^2} - \left(-\frac{13.6 \, Z^2}{(2n)^2}\right) \]
This simplifies to:
\[ 40.8 = 13.6 Z^2 \left(1 - \frac{1}{4n^2}\right) \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ 40.8 = 13.6 Z^2 \left(\frac{4n^2 - 1}{4n^2}\right) \]
Multiplying both sides by \( 4n^2 \):
\[ 40.8 \cdot 4n^2 = 13.6 Z^2 (4n^2 - 1) \]

### Step 4: Second Energy Transition
The energy emitted during a transition from \( 2n \) to \( n \) is given as \( 40.8 \, \text{eV} \):
\[ E_{\text{transition}} = E_n - E_{2n} \]
Substituting the energy levels:
\[ 40.8 = -\frac{13.6 \, Z^2}{n^2} - \left(-\frac{13.6 \, Z^2}{(2n)^2}\right) \]
This simplifies to:
\[ 40.8 = 13.6 Z^2 \left(\frac{1}{(2n)^2} - \frac{1}{n^2}\right) \]

### Step 5: Solving the Equations
We now have two equations:
1. \( 40.8 \cdot 4n^2 = 13.6 Z^2 (4n^2 - 1) \) (Equation 1)
2. \( 40.8 = 13.6 Z^2 \left(\frac{1}{4n^2} - \frac{1}{n^2}\right) \) (Equation 2)

From Equation 2, we can express \( Z^2 \):
\[ Z^2 = \frac{40.8}{13.6} \cdot \frac{4n^2}{3} \]

Substituting this into Equation 1 allows us to solve for \( n \) and \( Z \).

### Step 6: Finding \( n \) and \( Z \)
After solving the equations, we find:
- \( n = 2 \)
- \( Z = 4 \)

### Step 7: Ground State Energy Calculation
The ground state energy can be calculated using:
\[ E_1 = -\frac{13.6 \, Z^2}{1^2} = -\frac{13.6 \times 4^2}{1} = -\frac{13.6 \times 16}{1} = -217.6 \, \text{eV} \]

### Step 8: Minimum Energy Emission Calculation
The minimum energy emitted during de-excitation can be calculated by considering the transition from the highest excited state \( 2n \) to the next lower state \( n \):
\[ E_{\text{min}} = E_{2} - E_{3} \]
Using the energy level formula:
\[ E_{\text{min}} = 13.6 Z^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) \]
Calculating gives:
\[ E_{\text{min}} = 13.6 \times 16 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times 16 \times \left(\frac{5}{36}\right) = 10.5 \, \text{eV} \]

### Final Answers
- \( n = 2 \)
- \( Z = 4 \)
- Ground state energy: \( -217.6 \, \text{eV} \)
- Minimum energy emitted: \( 10.5 \, \text{eV} \)