The ionization energy of gaseous Na atoms is 495.5 kJ mol$$^{-1}$$. The lowest possible frequency of light that ionizes a sodium atom is (h = $$6.626 \times 10^{-34}$$ Js, N$$_A$$ = $$6.022 \times 10^{23}$$ mol$$^{-1}$$)

The ionization energy given is 495.5 kJ mol⁻¹ for gaseous Na atoms. This energy is required to remove an electron from one mole of sodium atoms. However, to find the lowest frequency of light that ionizes a single sodium atom, we need the ionization energy per atom.

First, convert the ionization energy from kJ mol⁻¹ to J mol⁻¹. Since 1 kJ = 1000 J, multiply by 1000:

$$495.5 \text{kJ mol}^{-1} = 495.5 \times 1000 \text{J mol}^{-1} = 495500 \text{J mol}^{-1}.$$

Now, to find the energy per atom, divide by Avogadro's number, $$N_A = 6.022 \times 10^{23}$$ mol⁻¹, which gives the number of atoms per mole:

$$\text{Energy per atom} = \frac{495500 \text{J mol}^{-1}}{6.022 \times 10^{23} \text{atoms mol}^{-1}}.$$

Simplify the expression:

$$\text{Energy per atom} = \frac{495500}{6.022 \times 10^{23}} \text{J atom}^{-1}.$$

Calculate the numerical value. First, express 495500 in scientific notation: $$495500 = 4.955 \times 10^5$$. Then:

$$\text{Energy per atom} = \frac{4.955 \times 10^5}{6.022 \times 10^{23}} = \frac{4.955}{6.022} \times 10^{5 - 23} = \frac{4.955}{6.022} \times 10^{-18}.$$

Divide 4.955 by 6.022:

$$\frac{4.955}{6.022} \approx 0.8225.$$

So,

$$\text{Energy per atom} \approx 0.8225 \times 10^{-18} \text{J atom}^{-1} = 8.225 \times 10^{-19} \text{J atom}^{-1}.$$

The lowest frequency of light that can ionize the atom corresponds to a photon with energy equal to the ionization energy per atom. The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency. Therefore,

$$\nu = \frac{E}{h}.$$

Given $$h = 6.626 \times 10^{-34}$$ J s, substitute the values:

$$\nu = \frac{8.225 \times 10^{-19}}{6.626 \times 10^{-34}} \text{s}^{-1}.$$

Simplify:

$$\nu = \frac{8.225}{6.626} \times 10^{-19 - (-34)} = \frac{8.225}{6.626} \times 10^{15}.$$

Divide 8.225 by 6.626:

$$\frac{8.225}{6.626} \approx 1.2413.$$

So,

$$\nu \approx 1.2413 \times 10^{15} \text{s}^{-1}.$$

Rounding to three significant figures, as in the options, gives $$1.24 \times 10^{15}$$ s⁻¹.

Comparing with the options:

A. $$1.24 \times 10^{15}$$ s⁻¹

B. $$7.50 \times 10^4$$ s⁻¹

C. $$4.76 \times 10^{14}$$ s⁻¹

D. $$3.15 \times 10^{15}$$ s⁻¹

The value $$1.24 \times 10^{15}$$ s⁻¹ matches option A.

Hence, the correct answer is Option A.