Question:

A monochromatic light source with a wavelength of $\lambda = 400$ nm is incident on a metal surface with a work function of $W = 2.5$ eV. The intensity of the light source is $I = 2.0 \times 10^{-4}$ W/m$^2$.

(a) Calculate the energy of one photon of the incident light in electron volts (eV).

(b) Determine the minimum frequency of the incident light needed to cause the photoelectric effect on this surface.

(c) If the incident light has a frequency of $5.0 \times 10^{14}$ Hz, calculate the kinetic energy, maximum speed, and wavelength of the ejected photoelectrons.

(d) Explain why increasing the intensity of the incident light source does not affect the energy of the ejected photoelectrons.

Answer:

(a) To calculate the energy of one photon, we can use the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.63 \times 10^{-34}$ J$\cdot$s) and $c$ is the speed of light ($3.0 \times 10^8$ m/s).

Substituting the given values: $E = \frac{(6.63 \times 10^{-34} \, \text{J} \cdot \text{s})(3.0 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}}$

$E = 4.97 \times 10^{-19}$ J

To convert this energy to electron volts (eV), we use the conversion factor: $1 \, \text{eV} = 1.60 \times 10^{-19}$ J.

Thus, the energy of one photon is:

$E = \frac{4.97 \times 10^{-19} \, \text{J}}{1.60 \times 10^{-19} \, \text{J/eV}}$

$E = 3.10$ eV

Therefore, the energy of one photon of the incident light is 3.10 eV.

(b) The minimum frequency ($f_{\text{min}}$) required to cause the photoelectric effect can be found using the equation: $E = hf_{\text{min}}$, where $E$ is the energy of one photon and $h$ is Planck's constant.

Substituting the known values: $f_{\text{min}} = \frac{E}{h} = \frac{3.10 \, \text{eV}}{6.63 \times 10^{-34} \, \text{J} \cdot \text{s}}$

$f_{\text{min}} \approx 4.68 \times 10^{14}$ Hz

Therefore, the minimum frequency of the incident light needed to cause the photoelectric effect is $4.68 \times 10^{14}$ Hz.

(c) To calculate the kinetic energy of the ejected photoelectrons, we can use the equation: $E_{\text{max}} = hf - W$, where $E_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $W$ is the work function of the metal.

Substituting the given values: $E_{\text{max}} = (6.63 \times 10^{-34} \, \text{J} \cdot \text{s})(5.0 \times 10^{14} \, \text{Hz}) - (2.5 \, \text{eV})(1.60 \times 10^{-19} \, \text{J/eV})$

$E_{\text{max}} \approx 3.96 \times 10^{-19}$ J

To convert this energy to electron volts (eV), we divide it by the conversion factor: $1 \, \text{eV} = 1.60 \times 10^{-19}$ J.

$E_{\text{max}} = \frac{3.96 \times 10^{-19} \, \text{J}}{1.60 \times 10^{-19} \, \text{J/eV}}$

$E_{\text{max}} \approx 2.47$ eV

The maximum kinetic energy of the ejected photoelectrons is approximately 2.47 eV.

The maximum speed ($v_{\text{max}}$) of the ejected photoelectrons can be determined using the equation: $E_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$, where $m$ is the mass of the electron.

Rearranging the equation to solve $v_{\text{max}}$, we have:

$v_{\text{max}} = \sqrt{\frac{2E_{\text{max}}}{m}}$

Substituting the known values: $v_{\text{max}} = \sqrt{\frac{2(2.47 \, \text{eV})(1.60 \times 10^{-19} \, \text{J/eV})}{9.11 \times 10^{-31} \, \text{kg}}}$

$v_{\text{max}} \approx 1.57 \times 10^6$ m/s

Therefore, the maximum speed of the ejected photoelectrons is approximately $1.57 \times 10^6$ m/s.

The wavelength ($\lambda_{\text{min}}$) of the ejected photoelectrons can be found using the equation: $\lambda_{\text{min}} = \frac{c}{f_{\text{min}}}$, where $c$ is the speed of light and $f_{\text{min}}$ is the minimum frequency of the incident light.

Substituting the given values: $\lambda_{\text{min}} = \frac{3.0 \times 10^8 \, \text{m/s}}{4.68 \times 10^{14} \, \text{Hz}}$

$\lambda_{\text{min}} \approx 6.41 \times 10^{-7}$ m

Therefore, the wavelength of the ejected photoelectrons is approximately $6.41 \times 10^{-7}$ m.

(d) Increasing the intensity of the incident light source does not affect the energy of the ejected photoelectrons because the energy of the ejected photoelectrons only depends on the frequency of the incident light and the work function of the metal surface. The intensity of the light source determines the number of photons incident on the metal surface per unit area per unit time, but it does not change the energy of an individual photon. Hence, increasing the intensity will only increase the number of photoelectrons emitted per unit time, without affecting their energy.