Hydrogen like atoms (atomic number `= Z`) in a sample are in excited state with principal quantum number n. The emission spectrum of the sample has `15` different lines. The second most energetic photon emitted by the sample has energy of `27.2 eV`. Find Z.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the number of lines in the emission spectrum
The number of spectral lines (N) emitted by an atom in an excited state can be calculated using the formula:
\[ N = \frac{n(n-1)}{2} \]
Given that \( N = 15 \), we can set up the equation:
\[ \frac{n(n-1)}{2} = 15 \]

### Step 2: Solve for n
Multiplying both sides by 2 gives:
\[ n(n-1) = 30 \]
Now, we can rearrange this into a quadratic equation:
\[ n^2 - n - 30 = 0 \]
Factoring the quadratic equation:
\[ (n - 6)(n + 5) = 0 \]
Thus, \( n = 6 \) (since n must be a positive integer).

### Step 3: Identify the transitions
The most energetic photon corresponds to the transition from the highest energy level (n=6) to the lowest (n=1). The second most energetic photon corresponds to the transition from n=6 to n=2.

### Step 4: Calculate the energy of the photon emitted during the transition
The energy of the emitted photon can be calculated using the formula:
\[ E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the transition from n=6 to n=2:
- \( n_1 = 2 \)
- \( n_2 = 6 \)

Substituting these values into the formula gives:
\[ E = 13.6 \, Z^2 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \]
Calculating the fractions:
\[ \frac{1}{2^2} = \frac{1}{4} \]
\[ \frac{1}{6^2} = \frac{1}{36} \]
Finding a common denominator (36):
\[ \frac{1}{4} = \frac{9}{36} \]
Thus:
\[ \frac{1}{2^2} - \frac{1}{6^2} = \frac{9}{36} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9} \]

### Step 5: Set up the equation with the given energy
We know the energy of the second most energetic photon is given as \( 27.2 \, eV \):
\[ 27.2 = 13.6 \, Z^2 \left( \frac{2}{9} \right) \]

### Step 6: Solve for Z
Rearranging the equation:
\[ Z^2 = \frac{27.2 \times 9}{13.6 \times 2} \]
Calculating the right-hand side:
\[ Z^2 = \frac{27.2 \times 9}{27.2} = 4.5 \]
Thus:
\[ Z^2 = 9 \]
Taking the square root:
\[ Z = 3 \]

### Final Answer
The atomic number \( Z \) is \( 3 \).

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