Correct Option: (B) $3\cdot20 \times 10^{14}\text{ Hz}$

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

The energy of a photon ($E$) is related to its frequency ($\nu$) by the Planck-Einstein relation:

$$E = h\nu$$

Given data:

• Energy ($E$) = $1\cdot326\text{ eV}$
• Conversion to Joules: $1\text{ eV} = 1\cdot6 \times 10^{-19}\text{ J}$
• Planck’s constant ($h$) = $6\cdot63 \times 10^{-34}\text{ J}\cdot\text{s}$

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

First, convert the energy into SI units (Joules):

$$E = 1\cdot326 \times 1\cdot6 \times 10^{-19}\text{ J} = 2\cdot1216 \times 10^{-19}\text{ J}$$

Rearranging the formula to solve for frequency ($\nu$):

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

Substituting the values:

$$\nu = \frac{2\cdot1216 \times 10^{-19}}{6\cdot63 \times 10^{-34}}$$ $$\nu = 0\cdot32 \times 10^{15}\text{ Hz}$$

Step 3 (Final Concluded Answer): [1/3 Mark]

Expressing the result in standard scientific notation:

$$\nu = 3\cdot20 \times 10^{14}\text{ Hz}$$

Thus, the correct option is (B).

💡 EL10 Concept Insight: Think of a photon as a tiny packet of energy; the higher the frequency of the light, the more "vibrations" it has per second, which directly translates into more energy packed inside that little bundle.