Given, Energy of the third energy level, $E_3 = 5.54 \times 10^{-10} \text{ erg}$ Energy of the second energy level, $E_2 = 2.44 \times 10^{-11} \text{ erg}$ The energy difference, $\Delta E$, is: $$ \begin{aligned} \Delta E &= E_3 - E_2 \\ &= 5.54 \times 10^{-10} \text{ erg} - 0.244 \times 10^{-10} \text{ erg} \\ &= 5.296 \times 10^{-10} \text{ erg} \end{aligned} $$ Converting the energy to Joules ($1 \text{ erg} = 10^{-7} \text{ J}$): $$ \begin{aligned} \Delta E &= 5.296 \times 10^{-10} \times 10^{-7} \text{ J} \\ &= 5.296 \times 10^{-17} \text{ J} \end{aligned} $$ Using the relation $\Delta E = h\nu$, where $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$: $$ \begin{aligned} u &= \frac{\Delta E}{h} \\ &= \frac{5.296 \times 10^{-17}}{6.626 \times 10^{-34}} \text{ Hz} \\ &= 7.99 imes 10^{16} \text{ Hz} \end{aligned} $$ The frequency of the emitted radiation is $7.99 \times 10^{16} \text{ Hz}$.