Atomic Structure

Definitions

Photon: elementary particle that is a quantum of electromagnetic radiation
Electromagentic (EM) Radiation: form of energy consisting of waves of electric and magnetic fields. EM radiations travels through space at the speed of light, 3 x 108 m/s

Energy Carried by a Photon

The energy carried by a photon can be described with the following equation:

\(\ce{E = h\nu}\)

Where E is the energy in joules, h is Planck's constant, and ν (Greek nu) is the frequency of the radiation (Hz, s-1).

Wavelength of Electromagnetic Wave
The wavelength of an electromagnetic (EM) wave can be determined as follows:

\(\ce{c = \nu\lambda}\)

\(\ce{\lambda = \frac{c}{\nu}}\)

Where c represents the speed of the light, ν represents the frequency, and λ represents the wavelength.

de Broglie Hypothesis

The wavelength of a particle based on Planck's constant (h) and momentum (p) is as follows:

\(\ce{\lambda = \frac{h}{p}}\)

Photoelectric Effect

The photoelectric effect occurs when electromagnetic radiation results in the emission of electrons from a material. The emitted electrons are known as photoelectrons. \[ K.E._{\text{max}} = h \nu - \phi \] The above equation states that the maximum kinetic energy of the ejected electrons is equal to the enrgy of the incoming photon minus the work function (φ) of the material. The work function is the minimum amount of energy required to remove an electron from the surface of a material.

Due to the conservation of energy, any excess energy remaining after the photon strikes the metal surface will be converted to kinetic energy in the ejected electron.

Consider the following problem:

A beam of light with a wavelength of 250 nm strikes a metal surface, causing electrons to be ejected from the surface. The work function (φ) of the metal is 2.0 eV. Calculate the velocity of the ejected electrons.

  1. Calculate the energy of the incident photons:

    The energy of a photon can be calculated using the equation: \[ E = \frac{hc}{\lambda} \] where:
    - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \)),
    - \( c \) is the speed of light (\( 3.00 \times 10^{8} \, \mathrm{m/s} \)),
    - \( \lambda \) is the wavelength of the light. Converting the wavelength from nm to meters: \[ \lambda = 250 \, \mathrm{nm} = 250 \times 10^{-9} \, \mathrm{m} \] Now, substituting the values into the equation: \[ E = \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s})(3.00 \times 10^{8} \, \mathrm{m/s})}{250 \times 10^{-9} \, \mathrm{m}} \] \[ E = 7.95 \times 10^{-19} \, \mathrm{J} \] Converting this energy into electron volts (1 eV = \( 1.602 \times 10^{-19} \) J): \[ E = \frac{7.95 \times 10^{-19} \, \mathrm{J}}{1.602 \times 10^{-19} \, \mathrm{J/eV}} \] \[ E \approx 4.96 \, \mathrm{eV} \]
  2. Calculate the kinetic energy of the ejected electrons:

    The kinetic energy (KE) of the ejected electrons is given by the energy of the incident photons minus the work function (\(\phi\)) of the metal: \[ \mathrm{KE} = E - \phi \] Substituting the values: \[ \mathrm{KE} = 4.96 \, \mathrm{eV} - 2.0 \, \mathrm{eV} \] \[ \mathrm{KE} = 2.96 \, \mathrm{eV} \] Converting this kinetic energy into joules: \[ \mathrm{KE} = 2.96 \times 1.602 \times 10^{-19} \, \mathrm{J} \] \[ \mathrm{KE} \approx 4.74 \times 10^{-19} \, \mathrm{J} \]
  3. Calculate the velocity of the ejected electrons:

    The kinetic energy of the electrons is also given by the equation: \[ \mathrm{KE} = \frac{1}{2}mv^2 \] where:
    - \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \mathrm{kg} \)),
    - \( v \) is the velocity of the electron. Rearranging to solve for \( v \): \[ v = \sqrt{\frac{2 \times \mathrm{KE}}{m}} \] Substituting the values: \[ v = \sqrt{\frac{2 \times 4.74 \times 10^{-19} \, \mathrm{J}}{9.11 \times 10^{-31} \, \mathrm{kg}}} \] \[ v = \sqrt{1.04 \times 10^{12} \, \mathrm{m^2/s^2}} \] \[ v \approx 1.02 \times 10^{6} \, \mathrm{m/s} \]
  4. Answer:

    The velocity of the ejected electrons is approximately \( 1.02 \times 10^6 \, \mathrm{m/s} \).