Question:

A monochromatic light with a wavelength of 500 nm is incident on a metal surface. The work function of this metal is 2 eV. The stopping potential required to stop the emitted photoelectrons is found to be 1.5 V. (a) Calculate the frequency of the incident light. (b) Determine the energy of each photon. (c) Calculate the maximum kinetic energy of the emitted photoelectrons. (d) Find the threshold frequency of the metal surface. (e) State and explain whether any electrons will be emitted from the metal surface if the intensity of the incident light is reduced. (f) Explain how the stopping potential is related to the maximum kinetic energy of the emitted photoelectrons.

Answer:

(a) To calculate the frequency of the incident light, we can use the relationship between the speed of light, wavelength, and frequency:

$$\text{Speed of light} = \text{Wavelength} \times \text{Frequency}$$

Rearranging the equation to solve for frequency:

$$\text{Frequency} = \frac{\text{Speed of light}}{\text{Wavelength}}$$

Given: [\text{Wavelength} = 500 , \text{nm} = 500 \times 10^{-9} , \text{m}] [\text{Speed of light} = 3 \times 10^8 , \text{m/s}]

Substituting the values into the equation:

$$\text{Frequency} = \frac{3 \times 10^8 \, \text{m/s}}{500 \times 10^{-9} \, \text{m}}$$

Calculating the frequency:

$$\text{Frequency} = 6 \times 10^{14} \, \text{Hz}$$

Therefore, the frequency of the incident light is $6 \times 10^{14}$ Hz.

(b) The energy of each photon can be calculated using the energy-frequency relationship:

$$E = h \times \text{Frequency}$$

where $E$ is the energy of the photon and $h$ is Planck's constant.

Given: [\text{Frequency} = 6 \times 10^{14} , \text{Hz}] $h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s}$

Substituting the values into the equation:

$$E = (6.63 \times 10^{-34} \, \text{J}\cdot\text{s}) \times (6 \times 10^{14} \, \text{Hz})$$

Calculating the energy of each photon:

$$E = 3.98 \times 10^{-19} \, \text{J}$$

Therefore, the energy of each photon is $3.98 \times 10^{-19}$ J.

(c) The maximum kinetic energy of the emitted photoelectrons can be calculated using the equation:

$$E_{\text{max}} = E - \phi$$

where $E_{\text{max}}$ is the maximum kinetic energy, $E$ is the energy of each photon, and $\phi$ is the work function of the metal.

Given: $E = 3.98 \times 10^{-19} \, \text{J}$ $\phi = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J}$

Substituting the values into the equation:

$$E_{\text{max}} = (3.98 \times 10^{-19} \, \text{J}) - (2 \times 1.6 \times 10^{-19} \, \text{J})$$

Calculating the maximum kinetic energy:

$$E_{\text{max}} = 7.8 \times 10^{-20} \, \text{J}$$

Therefore, the maximum kinetic energy of the emitted photoelectrons is $7.8 \times 10^{-20}$ J.

(d) The threshold frequency of the metal surface is the minimum frequency required to emit electrons. We can find it by equating the energy of a photon with the work function:

$$E = h \times \text{Threshold frequency}$$

Given: $E = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J}$ $h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s}$

Substituting the values into the equation:

$$2 \times 1.6 \times 10^{-19} \, \text{J} = (6.63 \times 10^{-34} \, \text{J}\cdot\text{s}) \times \text{Threshold frequency}$$

Solving for the threshold frequency:

$$\text{Threshold frequency} = \frac{2 \times 1.6 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J}\cdot\text{s}}$$

Calculating the threshold frequency:

$$\text{Threshold frequency} = 4.84 \times 10^{14} \, \text{Hz}$$

Therefore, the threshold frequency of the metal surface is $4.84 \times 10^{14}$ Hz.

(e) No electrons will be emitted from the metal surface if the intensity of the incident light is reduced. The photoelectric effect only depends on the energy of the incident photons, not their intensity. Reducing the intensity will decrease the number of photons incident on the metal surface, but it won't affect the energy of each photon. As long as each photon has energy greater than or equal to the work function, electrons will continue to be emitted.

(f) The stopping potential is related to the maximum kinetic energy of the emitted photoelectrons by the equation:

$$E_{\text{max}} = qV_{\text{stop}}$$

where $E_{\text{max}}$ is the maximum kinetic energy, $q$ is the charge of the electron ($1.6 \times 10^{-19}$ C), and $V_{\text{stop}}$ is the stopping potential.

Given: $E_{\text{max}} = 7.8 \times 10^{-20} \, \text{J}$ $q = 1.6 \times 10^{-19} \, \text{C}$

Substituting the values into the equation:

$$7.8 \times 10^{-20} \, \text{J} = (1.6 \times 10^{-19} \, \text{C}) \times V_{\text{stop}}$$

Solving for the stopping potential:

$$V_{\text{stop}} = \frac{7.8 \times 10^{-20} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}}$$

Calculating the stopping potential:

$$V_{\text{stop}} = 0.4875 \, \text{V}$$

Therefore, the stopping potential is approximately 0.4875 V. The maximum kinetic energy of the emitted photoelectrons is directly proportional to the stopping potential. As the stopping potential increases, the maximum kinetic energy also increases. This relationship is a direct consequence of the conservation of energy principle, where the energy of the photon is transferred to the electron in the form of maximum kinetic energy.