Question

A particular ultraviolet laser produces radiation of wavelength \(355 \mathrm{nm}\). Suppose this is used in a photoelectric experiment with a calcium sample. What will the stopping potential be?

Short answer

Answer: The stopping potential for the photoelectric experiment with the calcium sample is approximately 9.4 V.

Step-by-step solution

Calculate the frequency of the incoming radiation.

To find the frequency of the incoming radiation, we can use the relationship between the speed of light (\( c \)), wavelength (\(\lambda\)), and frequency (\(\nu\)): \( \nu = \frac{c}{\lambda} \) where \( c = 3 \times 10^8 \ m/s \) and \( \lambda = 355 \times 10^{-9} \ m \). Plugging in the values and calculating the frequency: \( \nu = \frac{3 \times 10^8 \ m/s}{355 \times 10^{-9} \ m} = 8.45 \times 10^{14} \ Hz \)

Find the work function of calcium.

The work function for calcium can be looked up from reference tables or online resources. The work function for calcium is approximately \(2.9 \ eV\).

Calculate the kinetic energy of the emitted electrons.

Using the photoelectric effect equation, we can find the kinetic energy of the emitted electrons: \( E_k = h \nu - \phi \) Here, \( h = 4.14 \times 10^{-15} \ eV \cdot s \) is the Planck's constant, \( \nu = 8.45 \times 10^{14} \ Hz \) is the frequency calculated in Step 1, and \( \phi = 2.9 \ eV \) is the work function of the calcium. Plugging in the values: \( E_k = (4.14 \times 10^{-15} \ eV \cdot s)(8.45 \times 10^{14} \ Hz) - 2.9 \ eV = 1.5 \ eV \)

Calculate the stopping potential.

Now we can calculate the stopping potential, \( V_s \), using the relationship between the kinetic energy and the stopping potential: \( E_k = eV_s \) Since the elementary charge \(e = 1.6 \times 10^{-19} \ C\), we can find the stopping potential: \( V_s = \frac{E_k}{e} = \frac{1.5 \ eV}{1.6 \times 10^{-19} \ C} = 9.4 \ V \) The stopping potential of the photoelectric experiment with the calcium sample will be about \(9.4 \ V\).

Key concepts

Stopping Potential

The stopping potential in the context of the photoelectric effect refers to the minimum voltage necessary to stop the most energetic photoelectrons emitted from a material due to incident light. When light with sufficient energy strikes a metal surface, it can eject electrons from the material, a phenomenon known as the photoelectric effect. If we apply a reverse voltage, we can determine how much potential is required to prevent these electrons from reaching the other side of the circuit—hence the term 'stopping potential.'

To compute the stopping potential, as seen in the original exercise, equation \( E_k = eV_s \) is used, where \( E_k \) is the kinetic energy of the photoelectrons, \( e \) is the elementary charge (approximately \(1.6 \times 10^{-19} C\)), and \( V_s \) is the stopping potential. By rearranging the formula \( V_s = \frac{E_k}{e} \), we can find the value of \( V_s \) that will halt the photoelectrons' motion. In our example, the calculated stopping potential was approximately 9.4 volts, signifying the energy conversion from the electrons' kinetic energy to electrical potential energy.

Work Function

The work function is a fundamental property of a material that represents the minimum energy required to remove an electron from the surface of a solid. Different materials have distinct work functions and this value is significant in the photoelectric effect. The energy of the incident photons must be higher than the work function for the photoelectric effect to occur. If the photon's energy is equal to the work function, the electron will just be able to escape the surface without any residual kinetic energy.

In our exercise, the work function \( \phi \) for calcium is given as approximately 2.9 eV (electron volts), indicating that each photon must supply at least this amount of energy to liberate an electron from calcium. Understanding the work function is crucial because it lets us calculate the kinetic energy imparted to the photoelectrons after they overcome this initial energy barrier.

Planck's Constant

Planck's constant, denoted as \( h \), is a vital quantity in quantum mechanics, acting as the proportionality constant between the energy \( E \) of a photon and its frequency \( u \), according to the relation \( E = h u \). Its value is approximately \(4.14 \times 10^{-15} eV \cdot s\) and is instrumental in calculating the energy of photons required to initiate the photoelectric effect.

This tiny constant reflects the quantized nature of energy at the microscopic level and its value is essential for a myriad calculations in quantum physics. Notably, it connects the macroscopic observation to the microscopic world in our exercise, allowing the calculation of the kinetic energy of photoelectrons based on the incident light's frequency.

Kinetic Energy of Photoelectrons

The kinetic energy of photoelectrons is the energy that these electrons have as they are ejected from a material due to the photoelectric effect. It is the excess energy of the incident photons beyond the work function of the material. The kinetic energy \( E_k \) of an electron is given by the equation \( E_k = h u - \phi \), where \( h \) is Planck's constant, \( u \) is the frequency of the incident light, and \( \phi \) is the work function.

In the context of our problem, after illuminating calcium with ultraviolet light, the photoelectrons that are ejected will possess kinetic energy depending on the frequency of the light and the work function of calcium. This kinetic energy dictates how fast the electrons are moving when ejected, which in turn informs us about the required stopping potential to halt these energetic electrons.