The problem involves the photoelectric effect. We are given the work function ($\phi$) of a metallic plate, the kinetic energy ($KE$) of the emitted photoelectrons, and Planck's constant ($h$). We need to find the angular frequency ($\omega$) of the incident light.

Photoelectric Effect Principle

According to Einstein's photoelectric equation, the energy of an incident photon ($E_{photon}$) is used to overcome the work function of the metal and provide kinetic energy to the emitted electron:

$E_{photon} = \phi + KE_{max}$

We are given:

• Work function, $\phi = 110 \times 10^{-20} \text{ J}$
• Maximum kinetic energy, $KE_{max} = 0 \text{ J}$ (since photoelectrons have zero kinetic energy)
• Planck's constant, $h = 6.63 \times 10^{-34} \text{ J.s}$

Substituting the values into the equation:

$E_{photon} = (110 \times 10^{-20} \text{ J}) + 0 \text{ J}$

$E_{photon} = 110 \times 10^{-20} \text{ J}$

Photon Energy and Angular Frequency

The energy of a photon can also be expressed in terms of its angular frequency ($\omega$) using Planck's constant:

$E_{photon} = \frac{h\omega}{2\pi}$

Angular Frequency Calculation

Now, we equate the two expressions for photon energy:

$\frac{h\omega}{2\pi} = \phi$

We need to solve for the angular frequency, $\omega$:

$\omega = \frac{2\pi\phi}{h}$

Substitute the given values:

$\omega = \frac{2\pi \times (110 \times 10^{-20} \text{ J})}{6.63 \times 10^{-34} \text{ J.s}}$

Using $\pi \approx 3.14159$:

$\omega = \frac{2 \times 3.14159 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}} \text{ rad/s}$

$\omega = \frac{691.15 \times 10^{-20}}{6.63 \times 10^{-34}} \text{ rad/s}$

$\omega \approx 104.25 \times 10^{(-20 - (-34))} \text{ rad/s}$

$\omega \approx 104.25 \times 10^{14} \text{ rad/s}$

Expressing this in scientific notation with three significant figures:

$\omega \approx 1.04 \times 10^{16} \text{ rad/s}$

Final Answer Derivation

The calculated angular frequency is approximately $1.04 \times 10^{16}$ rad/s. This matches Option D.