Step-by-step Solution:

1. Energy Levels of a Hydrogen-like Atom

   The energy of an electron in the $n$-th orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $E_n = -E_0 \frac{Z^2}{n^2}$ where $E_0 = 13.6 \text{ eV}$ is the ionization energy of the hydrogen atom.

2. Energy of an Emitted Photon

   When an electron makes a transition from a higher energy state $n_i$ to a lower energy state $n_f$, a photon is emitted. The energy of this photon, $\Delta E$, is the difference between the initial and final energy levels: $\Delta E = E_{n_i} - E_{n_f} = \left(-E_0 \frac{Z^2}{n_i^2}\right) - \left(-E_0 \frac{Z^2}{n_f^2}\right) = E_0 Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

3. Calculate Photon Energy for the $n=2 \to n=1$ Transition

   For this transition, $n_i=2$ and $n_f=1$. Let's denote the energy of the emitted photon as $\Delta E_{21}$. $\Delta E_{21} = E_0 Z^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = E_0 Z^2 \left(1 - \frac{1}{4}\right) = \frac{3}{4} E_0 Z^2$

4. Calculate Photon Energy for the $n=3 \to n=2$ Transition

   For this transition, $n_i=3$ and $n_f=2$. Let's denote the energy of the emitted photon as $\Delta E_{32}$. $\Delta E_{32} = E_0 Z^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = E_0 Z^2 \left(\frac{1}{4} - \frac{1}{9}\right) = E_0 Z^2 \left(\frac{9-4}{36}\right) = \frac{5}{36} E_0 Z^2$

5. Set up the Equation Based on the Problem Statement

   The problem states that the photon emitted in the $n=2 \to n=1$ transition has energy $74.8 \text{ eV}$ higher than the photon from the $n=3 \to n=2$ transition. Mathematically: $\Delta E_{21} = \Delta E_{32} + 74.8 \text{ eV}$ Rearranging this gives: $\Delta E_{21} - \Delta E_{32} = 74.8 \text{ eV}$

6. Solve for the Atomic Number Z

   Substitute the expressions for $\Delta E_{21}$ and $\Delta E_{32}$ into the equation: $\frac{3}{4} E_0 Z^2 - \frac{5}{36} E_0 Z^2 = 74.8$ Factor out $E_0 Z^2$: $E_0 Z^2 \left(\frac{3}{4} - \frac{5}{36}\right) = 74.8$ Find a common denominator for the fractions in the parenthesis: $E_0 Z^2 \left(\frac{27}{36} - \frac{5}{36}\right) = 74.8$ $E_0 Z^2 \left(\frac{22}{36}\right) = 74.8$ Substitute the value $E_0 = 13.6 \text{ eV}$: $13.6 \times Z^2 \times \frac{22}{36} = 74.8$ Now, solve for $Z^2$: $Z^2 = \frac{74.8 \times 36}{13.6 \times 22}$ We can simplify the calculation:

   • $74.8 / 13.6 = 5.5$
   • $22 / 5.5 = 4$, so $5.5 / 22 = 1/4$ So, the expression for $Z^2$ becomes: $Z^2 = \frac{5.5 \times 36}{22} = \frac{1}{4} \times 36 = 9$ Taking the square root to find $Z$: $Z = \sqrt{9} = 3$ Since the atomic number $Z$ must be a positive integer, the value of $Z$ is 3.

Final Answer

The value of $Z$ is 3.