The atom of a hydrogen' like gas are in a particulạr excited energy level. When these ATOMS de-excite, they 'emit photons of different energies. The maximum and minimum energies of emitted photons are `E_(max )=204 eV` and `E_(min )=10.57 eV`, respectively. The principal quantum number of initially. excited energy level is

Answer

To find the principal quantum number of the initially excited energy level of a hydrogen-like atom, we can follow these steps:

### Step 1: Understand the Energy Levels
In a hydrogen-like atom, the energy levels are given by the formula:
\[ E_n = -\frac{R_H Z^2}{n^2} \]
where \( R_H \) is the Rydberg constant for hydrogen (approximately 13.6 eV), \( Z \) is the atomic number, and \( n \) is the principal quantum number.

### Step 2: Determine the Maximum and Minimum Energies
The maximum energy photon emitted corresponds to the transition from the excited state \( n \) to the ground state \( n=1 \):
\[ E_{max} = E_n - E_1 = -\frac{R_H Z^2}{n^2} - \left(-\frac{R_H Z^2}{1^2}\right) = R_H Z^2 \left(1 - \frac{1}{n^2}\right) \]
Given \( E_{max} = 204 \, \text{eV} \), we have:
\[ R_H Z^2 \left(1 - \frac{1}{n^2}\right) = 204 \]

The minimum energy photon emitted corresponds to the transition from the excited state \( n \) to the state \( n-1 \):
\[ E_{min} = E_n - E_{n-1} = -\frac{R_H Z^2}{n^2} - \left(-\frac{R_H Z^2}{(n-1)^2}\right) = R_H Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) \]
Given \( E_{min} = 10.57 \, \text{eV} \), we have:
\[ R_H Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 10.57 \]

### Step 3: Set Up the Equations
Now we have two equations:
1. \( R_H Z^2 \left(1 - \frac{1}{n^2}\right) = 204 \) (Equation 1)
2. \( R_H Z^2 \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 10.57 \) (Equation 2)

### Step 4: Solve for \( Z^2 \)
From Equation 1:
\[ R_H Z^2 = \frac{204}{1 - \frac{1}{n^2}} \]

Substituting \( R_H Z^2 \) into Equation 2 gives:
\[ \frac{204}{1 - \frac{1}{n^2}} \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) = 10.57 \]

### Step 5: Simplify the Equation
The term \( \frac{1}{(n-1)^2} - \frac{1}{n^2} \) can be simplified:
\[ \frac{1}{(n-1)^2} - \frac{1}{n^2} = \frac{n^2 - (n-1)^2}{(n-1)^2 n^2} = \frac{n^2 - (n^2 - 2n + 1)}{(n-1)^2 n^2} = \frac{2n - 1}{(n-1)^2 n^2} \]

Substituting this back into the equation gives:
\[ \frac{204(2n - 1)}{(1 - \frac{1}{n^2})(n-1)^2 n^2} = 10.57 \]

### Step 6: Solve for \( n \)
Now we can rearrange and solve for \( n \). This may involve some algebraic manipulation and possibly numerical methods to find \( n \).

After solving, we find that \( n = 4 \).

### Final Answer
The principal quantum number of the initially excited energy level is:
\[ n = 4 \]

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