The emission frequency of radiation from a hydrogen-like ion during electron transitions can be understood using the formula derived from Rydberg's equation for hydrogen-like atoms, which is given as:

$$E = \frac{E_0}{h} \left( \frac{1}{n^2_1} - \frac{1}{n^2_2} \right)$$

where:

• $E$ is the energy of the emitted photon,
• $E_0$ is the Rydberg constant for hydrogen,
• $h$ is Planck's constant,
• $n_1$ and $n_2$ are the principal quantum numbers for the initial and final energy levels, respectively, with $n_2 < n_1$.

Given a transition from $n=2$ to $n=1$ emits radiation with a frequency of $3 \times 10^{15} \mathrm{~Hz}$, we can write:

$$\nu_1 = 3 \times 10^{15} \mathrm{~Hz}$$

For the transition from $n=3$ to $n=1$, we can use the same principle to find the frequency of the emitted radiation, $\nu_2$, which will depend on the difference in energy levels involved in the transition. The frequency is directly proportional to this energy difference, so we can compare the two transitions using their respective frequencies:

$$\frac{\nu_2}{\nu_1} = \frac{\left( \frac{1}{n_{1f}^2} - \frac{1}{n_{1i}^2} \right)}{\left( \frac{1}{n_{2f}^2} - \frac{1}{n_{2i}^2} \right)} = \frac{\left( \frac{1}{1^2} - \frac{1}{3^2} \right)}{\left( \frac{1}{1^2} - \frac{1}{2^2} \right)} = \frac{\left(1 - \frac{1}{9}\right)}{\left(1 - \frac{1}{4}\right)}$$

After simplification:

$$\frac{\nu_2}{3 \times 10^{15}} = \frac{\left(\frac{8}{9}\right)}{\left(\frac{3}{4}\right)} = \frac{32}{27}$$

Thus, to find the frequency $ u_2$ for the transition from $n=3$ to $n=1$:

$$\nu_2 = \frac{32}{9} \times 10^{15} \mathrm{Hz}$$

Hence, in the given formula $\frac{x}{9} \times 10^{15} \mathrm{~Hz}$ for the frequency, $x$ is equal to $32$.