Modern Physics - Quantum Mechanics - SS2 Physics Past Questions and Answers - page 2

To calculate the kinetic energy of the emitted electrons, we first need to determine if the incident light has enough energy to overcome the work function of the metal. If the frequency of the light is greater than or equal to the threshold frequency, it can emit electrons. In this case, the frequency (6.0 × 1014 Hz) is greater than the threshold frequency (5.0 × 1014 Hz), so electrons will be emitted.

The energy of a photon can be calculated using the equation: E = hf, where E is the energy, h is Planck's constant (6.63 × 10-34 J·s), and f is the frequency.

E = (6.63 × 10-34 J·s) × (6.0 × 1014 Hz) = 3.98 × 10-19 J

The kinetic energy of the emitted electrons can be calculated by subtracting the work function from the energy of the incident photons:

Kinetic energy = E - Work function

Kinetic energy = (3.98 × 10-19 J) - (4.0 eV × 1.6 × 10-19 J/eV) 

= -2.88 × 10-20 J

Note: The negative sign indicates that the electrons have lost energy, which is consistent with the emission of electrons in the photoelectric effect.

To calculate the maximum kinetic energy of the emitted electrons, we need to convert the given wavelength to frequency using the formula: f = c / λ, where f is the frequency, c is the speed of light (3.0 × 108 m/s), and λ is the wavelength.

f = (3.0 × 108 m/s) / (500 × 10-9 m) = 6.0 × 1014 Hz

Now, we can calculate the energy of the incident photons using the formula: E = hf.

E = (6.63 × 10-34 J·s) × (6.0 × 1014 Hz) = 3.98 × 10-19 J

The maximum kinetic energy of the emitted electrons can be calculated by subtracting the work function from the energy of the incident photons:

Kinetic energy = E - Work function

Kinetic energy = (3.98 × 10-19 J) - (2.0 eV × 1.6 × 10-19 J/eV) 

= 1.38 × 10-19 J