A certain species of ionized ATOMS (hydrogen like) produces an emission line spectrum according to Bohr model. A group of lines in the spectrum form a series in which the smatlest energy is `4.896 eV` and the maximum energy is `13.6 eV`. The atomic number of the atom is

Answer

To find the atomic number of the hydrogen-like ionized atom based on the given emission line spectrum energies, we can follow these steps:

### Step 1: Understand the Energy Levels
In the Bohr model, the energy levels of a hydrogen-like atom are given by the formula:

\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]

where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 2: Identify the Maximum and Minimum Energies
From the problem, we know:
- The maximum energy \( E_{\text{max}} = 13.6 \, \text{eV} \) corresponds to the ionization energy (when \( n \to \infty \)).
- The minimum energy \( E_{\text{min}} = 4.896 \, \text{eV} \) corresponds to a transition from a higher energy level to a lower energy level.

### Step 3: Determine the Energy Levels
For the maximum energy (ionization energy), we can set \( n = 1 \):
\[ E_1 = -\frac{Z^2 \cdot 13.6}{1^2} = -13.6 \, \text{eV} \]

For the minimum energy, we need to find the principal quantum number \( n \) for which:
\[ E_n = -\frac{Z^2 \cdot 13.6}{n^2} = 4.896 \, \text{eV} \]

### Step 4: Solve for \( Z \)
Since the energy levels are negative, we can express the energy difference as:
\[ E_{\text{transition}} = E_{n} - E_{1} = 4.896 - (-13.6) = 4.896 + 13.6 = 18.496 \, \text{eV} \]

This energy corresponds to a transition from \( n = 2 \) to \( n = 1 \):
\[ E_2 = -\frac{Z^2 \cdot 13.6}{2^2} = -\frac{Z^2 \cdot 13.6}{4} \]

Setting the equation:
\[ E_2 - E_1 = 18.496 \, \text{eV} \]
\[ -\frac{Z^2 \cdot 13.6}{4} + 13.6 = 18.496 \]

Rearranging gives:
\[ -\frac{Z^2 \cdot 13.6}{4} = 18.496 - 13.6 \]
\[ -\frac{Z^2 \cdot 13.6}{4} = 4.896 \]

### Step 5: Solve for \( Z^2 \)
Multiplying both sides by -4:
\[ Z^2 \cdot 13.6 = -4 \cdot 4.896 \]
\[ Z^2 \cdot 13.6 = -19.584 \]

Now divide by 13.6:
\[ Z^2 = \frac{-19.584}{-13.6} = 1.44 \]

Taking the square root gives:
\[ Z = \sqrt{1.44} = 1.2 \]

Since \( Z \) must be a whole number, we round to the nearest whole number, which is \( Z = 1 \).

### Final Answer
The atomic number of the atom is \( Z = 1 \).

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