C Step 1: Identify the formula for the energy of a photon. The energy of an emitted photon ($E$) is given by the formula: $$E = \frac{hc}{\lambda}$$ where $h$ is Planck's constant, $c$ is the speed of light (emitted radiation speed), and $\lambda$ is the wavelength of the photon. Step 2: Substitute the given values into the formula. Given: Planck's constant, $h = 6.63 \times 10^{-34} \text{ Js}$ Speed of emitted radiation, $c = 3.0 \times 10^8 \text{ ms}^{-1}$ Wavelength of emitted photon, $\lambda = 5.68 \times 10^{-6} \text{ m}$ $$E = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.0 \times 10^8 \text{ ms}^{-1})}{5.68 \times 10^{-6} \text{ m}}$$ Step 3: Calculate the energy. First, calculate the numerator: $$6.63 \times 3.0 = 19.89$$ $$10^{-34} \times 10^8 = 10^{(-34+8)} = 10^{-26}$$ So, the numerator is $19.89 \times 10^{-26} \text{ Jm}$. Now, divide by the wavelength: $$E = \frac{19.89 \times 10^{-26} \text{ Jm}}{5.68 \times 10^{-6} \text{ m}}$$ $$E = \left(\frac{19.89}{5.68}\right) \times \left(\frac{10^{-26}}{10^{-6}}\right) \text{ J}$$ $$E \approx 3.50176 \times 10^{(-26 - (-6))} \text{ J}$$ $$E \approx 3.50176 \times 10^{-20} \text{ J}$$ Step 4: Round the result and compare with the given options. Rounding to two decimal places for the coefficient, we get $3.50 \times 10^{-20} \text{ J}$. Comparing this to the options: A $1.49 \times 10^{-20} \text{ J}$ B $1.49 \times 10^{-19} \text{ J}$ C $3.49 \times 10^{-20} \text{ J}$ D $3.49 \times 10^{-19} \text{ J}$ The calculated value $3.50 \times 10^{-20} \text{ J}$ is closest to option C, $3.49 \times 10^{-20} \text{ J}$. The final answer is $\boxed{\text{C}}$.