Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_(n) = - (13.6 Z^(2))/(n^(2))`
When this excited makes a transition from excited state to ground state , most energetic photons have energy
`E_(max) = 52.224 eV`. and least energetic photons have energy
`E_(min) = 1.224 eV`
Find the atomic number of atom and the initial state or excitation.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the energy equations
The energy of a hydrogen-like atom in the nth excited state is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 2: Determine the energy difference for maximum energy transition
The maximum energy photon emitted during the transition from the nth excited state to the ground state (n=1) is given as:
\[ E_{max} = E_n - E_1 \]
Substituting the energy values, we have:
\[ E_{max} = \left(-\frac{13.6 Z^2}{n^2}\right) - \left(-\frac{13.6 Z^2}{1^2}\right) = 13.6 Z^2 \left(1 - \frac{1}{n^2}\right) \]
Setting this equal to the given maximum energy:
\[ 13.6 Z^2 \left(1 - \frac{1}{n^2}\right) = 52.224 \quad \text{(Equation 1)} \]

### Step 3: Determine the energy difference for minimum energy transition
The minimum energy photon emitted occurs when the transition is from the nth excited state to the (n-1)th state:
\[ E_{min} = E_n - E_{n-1} \]
Substituting the energy values, we have:
\[ E_{min} = \left(-\frac{13.6 Z^2}{n^2}\right) - \left(-\frac{13.6 Z^2}{(n-1)^2}\right) = 13.6 Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) \]
Setting this equal to the given minimum energy:
\[ 13.6 Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 1.224 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 1:
\[ 13.6 Z^2 \left(1 - \frac{1}{n^2}\right) = 52.224 \]
Rearranging gives:
\[ Z^2 \left(1 - \frac{1}{n^2}\right) = \frac{52.224}{13.6} \]
Calculating the right side:
\[ Z^2 \left(1 - \frac{1}{n^2}\right) = 3.84 \quad \text{(Equation 3)} \]

From Equation 2:
\[ 13.6 Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 1.224 \]
Rearranging gives:
\[ Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = \frac{1.224}{13.6} \]
Calculating the right side:
\[ Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 0.09 \quad \text{(Equation 4)} \]

### Step 5: Divide Equation 3 by Equation 4
Dividing Equation 3 by Equation 4:
\[ \frac{Z^2 \left(1 - \frac{1}{n^2}\right)}{Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right)} = \frac{3.84}{0.09} \]
This simplifies to:
\[ \frac{1 - \frac{1}{n^2}}{\frac{1}{(n-1)^2} - \frac{1}{n^2}} = 42.67 \]

### Step 6: Solve for \( n \)
Using trial and error, we can substitute values for \( n \) to find a suitable integer. Testing \( n = 5 \):
\[ \frac{1 - \frac{1}{25}}{\frac{1}{16} - \frac{1}{25}} = \frac{0.96}{0.09} = 42.67 \]
Thus, \( n = 5 \).

### Step 7: Find the atomic number \( Z \)
Substituting \( n = 5 \) back into Equation 3:
\[ Z^2 \left(1 - \frac{1}{25}\right) = 3.84 \]
This simplifies to:
\[ Z^2 \cdot \frac{24}{25} = 3.84 \]
Solving for \( Z^2 \):
\[ Z^2 = \frac{3.84 \cdot 25}{24} = 4 \]
Thus, \( Z = 2 \).

### Step 8: Conclusion
The atomic number of the atom is \( Z = 2 \) and the initial state of excitation is \( n = 5 \).

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