For a monochromatic beam, the associated electric field becomes zero 1.2 × 10^ 15 times per second. Determine the maximum kinetic energy of the photoelectrons emitted when this light is incident on a metal surface

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 0.48 eV

The electric field of an electromagnetic wave becomes zero twice in each cycle.

Frequency of the light,

\nu = \dfrac{1.2 \times 10^{15}}{2} = 6 \times 10^{14} \text{ Hz}

Energy of the incident photon,

E = h\nu

E = (4.14 \times 10^{-15} \text{ eV s}) \times (6 \times 10^{14} \text{ Hz}) = 2.484 \text{ eV}

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by:

K_\max = E - \Phi

K_\max = 2.484 \text{ eV} - 2.0 \text{ eV} = 0.484 \text{ eV} \approx 0.48 \text{ eV}

For a monochromatic beam, the associated electric field becomes...

Options

Correct answer

Step-by-step solution

More from Photoelectric Effect and Wave–Particle Duality