A hydrogen like atom of atomic 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 to 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(eV) that can be emitted by this atom during de - excitation. Ground state energy of hydrogen atom is -13.6 eV

Explanation

To solve the problem step by step, we will break down the information provided and use the relevant equations from quantum mechanics and the Bohr model of the hydrogen-like atom.

### Step 1: Understand the given information
- The atom is hydrogen-like with atomic number \( Z \).
- It is in an excited state with quantum number \( 2n \).
- The maximum energy photon emitted is \( 204 \, \text{eV} \).
- When it transitions to quantum state \( n \), a photon of energy \( 40.8 \, \text{eV} \) is emitted.
- The ground state energy of hydrogen atom is \( -13.6 \, \text{eV} \).

### Step 2: Use the energy formula for hydrogen-like atoms
The energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV} \]

### Step 3: Set up the equations based on the transitions
1. For the transition from \( 2n \) to \( 1 \) (maximum energy):
\[ E_{2n} - E_1 = 204 \, \text{eV} \]
Substituting the energy formula:
\[ \left(-\frac{13.6 Z^2}{(2n)^2}\right) - \left(-\frac{13.6 Z^2}{1^2}\right) = 204 \]
Simplifying gives:
\[ -\frac{13.6 Z^2}{4n^2} + 13.6 Z^2 = 204 \]
\[ 13.6 Z^2 \left(1 - \frac{1}{4n^2}\right) = 204 \]

2. For the transition from \( 2n \) to \( n \):
\[ E_{2n} - E_n = 40.8 \, \text{eV} \]
Substituting the energy formula:
\[ \left(-\frac{13.6 Z^2}{(2n)^2}\right) - \left(-\frac{13.6 Z^2}{n^2}\right) = 40.8 \]
Simplifying gives:
\[ -\frac{13.6 Z^2}{4n^2} + \frac{13.6 Z^2}{n^2} = 40.8 \]
\[ 13.6 Z^2 \left(\frac{1}{n^2} - \frac{1}{4n^2}\right) = 40.8 \]
\[ 13.6 Z^2 \left(\frac{3}{4n^2}\right) = 40.8 \]

### Step 4: Solve the equations
From the first equation:
\[ 13.6 Z^2 \left(1 - \frac{1}{4n^2}\right) = 204 \quad \text{(1)} \]
From the second equation:
\[ 13.6 Z^2 \left(\frac{3}{4n^2}\right) = 40.8 \quad \text{(2)} \]

#### Divide equation (1) by equation (2):
\[ \frac{1 - \frac{1}{4n^2}}{\frac{3}{4n^2}} = \frac{204}{40.8} \]
Calculating the right side:
\[ \frac{204}{40.8} = 5 \]
Thus:
\[ \frac{4n^2 - 1}{3} = 5 \]
\[ 4n^2 - 1 = 15 \]
\[ 4n^2 = 16 \quad \Rightarrow \quad n^2 = 4 \quad \Rightarrow \quad n = 2 \]

### Step 5: Substitute \( n \) back to find \( Z \)
Substituting \( n = 2 \) into equation (1):
\[ 13.6 Z^2 \left(1 - \frac{1}{4 \cdot 4}\right) = 204 \]
\[ 13.6 Z^2 \left(1 - \frac{1}{16}\right) = 204 \]
\[ 13.6 Z^2 \left(\frac{15}{16}\right) = 204 \]
\[ Z^2 = \frac{204 \cdot 16}{13.6 \cdot 15} \]
Calculating \( Z^2 \):
\[ Z^2 = \frac{3264}{204} = 16 \quad \Rightarrow \quad Z = 4 \]

### Step 6: Calculate the ground state energy
Using \( Z = 4 \) and \( n = 1 \):
\[ E_1 = -\frac{13.6 \cdot 4^2}{1^2} = -\frac{13.6 \cdot 16}{1} = -217.6 \, \text{eV} \]

### Step 7: Calculate the minimum energy emitted during de-excitation
The minimum energy emitted occurs when the electron transitions from \( 2n \) to \( 2n-1 \) (i.e., from \( 4 \) to \( 3 \)):
\[ E_4 - E_3 = \left(-\frac{13.6 \cdot 4^2}{16}\right) - \left(-\frac{13.6 \cdot 4^2}{9}\right) \]
Calculating:
\[ E_4 = -\frac{217.6}{16} = -13.6 \, \text{eV}, \quad E_3 = -\frac{217.6}{9} \approx -24.18 \, \text{eV} \]
Thus:
\[ E_{\text{min}} = E_4 - E_3 = -13.6 + 24.18 \approx 10.58 \, \text{eV} \]

### Final Answers:
- \( n = 2 \)
- \( Z = 4 \)
- Ground state energy \( = -217.6 \, \text{eV} \)
- Minimum energy emitted during de-excitation \( \approx 10.58 \, \text{eV} \)