Question:

A beam of light with a wavelength of 450 nm is incident on a metal surface. The work function of the metal is 2.25 eV. When the intensity of the light is increased, no change is observed in the maximum kinetic energy of the emitted electrons, but the number of emitted electrons per second increases.

a) Explain why increasing the intensity of the light does not affect the maximum kinetic energy of the emitted electrons.

b) Calculate the energy of a single photon of the incident light.

c) Calculate the frequency of the incident light.

d) Calculate the velocity of the fastest emitted electron.

Answer:

a) The photoelectric effect is a phenomenon in which electrons are ejected from the surface of a metal when light is incident on it. According to the wave theory of light, increasing the intensity of light is equivalent to increasing the amplitude of the electromagnetic wave. However, the photoelectric effect can only be explained by considering light as a particle-like entity known as the photon. The energy of a photon is directly proportional to its frequency, and increasing the intensity of light only increases the number of photons incident on the surface, not the energy of individual photons. The maximum kinetic energy of the emitted electrons is determined by the difference between the energy of a single photon (related to its frequency) and the work function of the metal, not the intensity of light. Therefore, increasing the intensity of the light does not affect the maximum kinetic energy of the emitted electrons.

b) The energy of a photon is given by the equation:

E = hf

Where E is the energy of the photon, h is the Planck's constant (6.63 x 10^-34 J·s), and f is the frequency of the photon. To calculate the energy of a single photon, we need to convert the wavelength of the light into frequency.

First, we use the equation:

c = λf

Where c is the speed of light (3 x 10^8 m/s), λ is the wavelength, and f is the frequency. Rearranging the equation to solve for f:

f = c/λ

Plugging in the given wavelength of 450 nm (1 nm = 1 x 10^-9 m):

f = (3 x 10^8 m/s) / (450 x 10^-9 m)

f = 6.67 x 10^14 Hz

Now, substitute this value of frequency into the equation for energy of a photon:

E = hf

E = (6.63 x 10^-34 J·s) × (6.67 x 10^14 Hz)

E = 4.42 x 10^-19 J

Therefore, the energy of a single photon of the incident light is 4.42 x 10^-19 J.

c) The frequency of the incident light is already calculated as 6.67 x 10^14 Hz.

d) The maximum velocity of the emitted electrons can be calculated using the equation:

KEmax = (1/2)mv^2

KEmax = Energy of incident photon - Work function

Since we are interested in the fastest emitted electron, all the energy of the incident photon is used to remove an electron from the surface of the metal.

Therefore,

KEmax = Energy of incident photon = 4.42 x 10^-19 J

To calculate the velocity, we can use the equation:

KEmax = (1/2)mv^2

Rearranging this equation to solve for v:

v = sqrt((2KEmax) / m)

The mass of an electron (m) is 9.11 x 10^-31 kg.

Substituting the values:

v = sqrt((2 × 4.42 x 10^-19 J) / (9.11 x 10^-31 kg))

v = 6.58 x 10^5 m/s

Therefore, the velocity of the fastest emitted electron is 6.58 x 10^5 m/s.