Question

The photoelectric work function of potassium is 2.3 eV. If light that has a wavelength of 190 nm falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

Short answer

(a) 4.26 V; (b) 4.26 eV; (c) \(9.73 \times 10^5\) m/s.

Step-by-step solution

Convert Wavelength to Energy

First, we need to convert the wavelength of the incoming light to energy using the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \) J\cdot s is the Planck's constant, \( c = 3 \times 10^8 \) m/s is the speed of light, and \( \lambda = 190 \) nm is the wavelength. Convert the wavelength from nanometers to meters by multiplying by \( 10^{-9} \). Substitute the values:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{190 \times 10^{-9}} \] This gives the energy in joules. To convert this energy to electron volts (eV), divide by the charge of an electron (~\(1.6 \times 10^{-19}\) C).

Calculate Energy in Electron Volts

Calculate the energy (\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{190 \times 10^{-9}} = 1.05 \times 10^{-18} \text{ J} \] Convert energy from joules to electron volts (1 eV = \(1.6 \times 10^{-19} \) J):\[ E = \frac{1.05 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 6.56 \text{ eV} \] This is the energy of the incoming photons in eV.

Determine the Kinetic Energy of Electrons

Using the photoelectric equation \( KE_{max} = E - \phi \), where \( E \) is the energy of the photons and \( \phi \) is the work function in eV (2.3 eV for potassium), we can find the maximum kinetic energy of the ejected electrons:\[ KE_{max} = 6.56 - 2.3 = 4.26 \text{ eV} \] This is the kinetic energy of the most energetic electrons ejected.

Calculate the Stopping Potential

The stopping potential \( V_s \) is simply the maximum kinetic energy divided by the electron charge to ensure electrons stop moving, given by:\[ V_s = KE_{max} \] Substitute the values:\[ V_s = 4.26 \text{ V} \] This is the stopping potential in volts.

Find the Speed of the Ejected Electrons

The kinetic energy \( KE \) is also given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the electron mass \(9.11 \times 10^{-31}\) kg and \( v \) is the speed. Solving for \( v \):\[ v = \sqrt{\frac{2 \times KE}{m}} \] Convert the kinetic energy back to joules (4.26 eV, so \( 4.26 \times 1.6 \times 10^{-19} \) J). Substitute in \( v = \sqrt{\frac{2 \times 4.26 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}}} \).Calculate \( v \) to find the speed of these electrons.

Key concepts

Work Function

The work function is a fundamental concept in the study of the photoelectric effect. It refers to the minimum energy required to remove an electron from the surface of a material. In the context of the exercise, potassium has a work function of 2.3 eV, meaning that any incident photon must have at least this amount of energy to eject an electron.

Understanding the work function is crucial because it sets the threshold for the photoelectric effect to occur. When light hits the surface of a metal, like potassium in this case, only photons with energy equal to or greater than the work function can liberate electrons. If the photon's energy is less, no electrons will be emitted, and the photoelectric effect cannot be observed.

The work function can vary from one material to another, making some materials more susceptible to the photoelectric effect than others.

Kinetic Energy of Electrons

When light strikes a metal surface and electrons are ejected due to the photoelectric effect, they gain kinetic energy. The kinetic energy (KE) of these electrons is what allows us to measure their speed and movement. In the given exercise, the energy of the incoming photon was calculated to be 6.56 eV, which is greater than the work function of 2.3 eV.

The formula to find the maximum kinetic energy of the electrons is:

• \( KE_{max} = E - \phi \)

Here, \( E \) is the energy of the photons (6.56 eV in this case) and \( \phi \) is the work function (2.3 eV). By substituting these values, we find that the maximum kinetic energy of the ejected electrons is 4.26 eV.

This calculation is important as it tells us about the excess energy the electrons have after overcoming the work function needed to escape the metal surface. This energy contributes to their velocity as they leave the surface.

Stopping Potential

The stopping potential, denoted as \( V_s \), is a critical concept used to measure the kinetic energy of ejected electrons in the photoelectric effect. It is defined as the electric potential needed to stop the most energetic photoelectrons from reaching the opposite electrode.

In simpler terms, it's the voltage needed to bring the fastest ejected electrons to a halt. In the given problem, since the maximum kinetic energy of electrons was determined to be 4.26 eV, directly the stopping potential required is also 4.26 V, since 1 eV corresponds to 1 V.

By applying this stopping potential, we create a reverse field that precisely cancels out the electrons' kinetic energy, effectively stopping their motion. Stopping potential experiments are instrumental in confirming Einstein's photoelectric equation and understanding the energy dynamics of photoelectrons.

Planck's Constant

Planck's constant, represented by the symbol \( h \), is a fundamental constant in quantum mechanics. It quantifies the relationship between the energy of a photon and the frequency of the electromagnetic wave from which the photon originates. The constant is approximately equal to \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\).

In the exercise, Planck's constant plays a pivotal role in determining the energy contained in photons of a particular wavelength. Using the equation:

• \( E = \frac{hc}{\lambda} \)

We see that the energy \( E \) of a photon is directly proportional to Planck's constant \( h \) and inversely proportional to the wavelength \( \lambda \). Here \( c \) represents the speed of light.

Thus, Planck's constant is the fee for nature's exchange between wave-like and particle-like behaviors of light. Its utilization in calculating photon energy allows us to bridge the gap between macroscopic electromagnetic waves and their microscopic photoelectric effects.