Problem 28 - University Physics with Modern Physics

Question

A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

Step-by-Step Solution

Answer

(a) 5.78 eV, (b) 1.56 × 10^18 electrons/second, (c) 7.8 × 10^17 electrons/second, (d) 7.8 × 10^17 electrons/second.

Step 1: Calculate the Energy of a Single Photon

The energy of a photon can be calculated using the formula: \[E = \frac{hc}{\lambda}\,,\]where \(h\) is Planck's constant \(6.626 \times 10^{-34}\, \text{J·s}\), \(c\) is the speed of light \(3 \times 10^{8}\, \text{m/s}\), and \(\lambda\) is the wavelength \(124\, \text{nm} = 124 \times 10^{-9}\, \text{m}\). Substituting these values, we find the photon energy:\[E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{124 \times 10^{-9}} = 1.601 \times 10^{-18} \text{J}.\]Convert the energy from joules to electron volts using \(1 \text{J} = 6.242 \times 10^{18} \text{eV}\):\[E = 1.601 \times 10^{-18} \times 6.242 \times 10^{18} = 9.94 \, \text{eV}.\]

Step 2: Calculate the Work Function of the Metal

The maximum kinetic energy \(K_{\text{max}}\) is given as \(4.16\, \text{eV}\). The work function \(\phi\) can be calculated using the photoelectric equation:\[K_{\text{max}} = E - \phi\]Substituting the known values,\[4.16 \, \text{eV} = 9.94 \, \text{eV} - \phi\]Solving for \(\phi\),\[\phi = 9.94 \, \text{eV} - 4.16 \, \text{eV} = 5.78 \, \text{eV}.\]

Step 3: Calculate the Number of Photoelectrons Ejected Each Second

The power of the beam is \(2.50\, \text{W}\), which is equivalent to \(2.50 \, \text{J/s}\). Each photon has an energy of \(1.601 \times 10^{-18} \, \text{J}\). The number of photons (and hence electrons) ejected per second is given by:\[\text{Number of photoelectrons} = \frac{2.50 \, \text{J/s}}{1.601 \times 10^{-18} \, \text{J/photon}}\]Calculating this gives approximately:\[1.56 \times 10^{18} \, \text{photoelectrons/second}.\]

Step 4: Effect of Halving the Power on Photoelectron Emission

If the power is reduced by half (i.e., \(1.25\, \text{W}\)), keeping the same photon energy, the number of photoelectrons per second will also be halved:\[\text{Number of photoelectrons} = \frac{1.25 \, \text{J/s}}{1.601 \times 10^{-18} \, \text{J/photon}} \approx 7.8 \times 10^{17} \, \text{photoelectrons/second}.\]

Step 5: Effect of Halving the Wavelength on Photoelectron Emission

If the wavelength is halved, \(\lambda = 62 \, \text{nm} = 62 \times 10^{-9} \, \text{m}\), the energy of each photon becomes:\[E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{62 \times 10^{-9}} = 3.202 \times 10^{-18} \, \text{J} \approx 19.88 \, \text{eV}\]Since power does not change, the rate of photoelectrons emission becomes:\[\text{Number of photoelectrons} = \frac{2.50 \, \text{J/s}}{3.202 \times 10^{-18} \, \text{J/photon}} \approx 7.8 \times 10^{17} \, \text{photoelectrons/second}\]Thus, the number of photoelectrons per second decreases.

Key Concepts

Photon Energy

Photon energy is a crucial aspect of understanding the photoelectric effect. When a beam of light, like the 2.50-W beam mentioned in the exercise, strikes a metal, it's made up of particles called photons. Each photon carries a specific amount of energy which can be calculated using the formula:

• \( E = \frac{hc}{\lambda} \)

Here, \(h\) is Planck's constant, \(6.626 \times 10^{-34} \, \text{J·s}\), \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light.
For instance, using a wavelength of 124 nanometers, we find that each photon has an energy of \(1.601 \times 10^{-18}\) joules—or approximately 9.94 electron volts (eV) when converted.
This energy represents the ability of the photon to accomplish work, like knocking electrons free from a metal surface.

Work Function

The work function is the minimum amount of energy needed to eject an electron from the surface of a metal. It's a critical parameter because it determines how easily electrons can be freed by photons from a particular metal.
Mathematically, the work function \( \phi \) can be determined using the photoelectric equation:

• \( K_{\text{max}} = E - \phi \)

In this exercise, the maximum kinetic energy \( K_{\text{max}} \) of the ejected electrons was given as 4.16 eV. We already calculated the photon energy to be 9.94 eV. Using these values, the work function of the metal calculates to 5.78 eV.
This difference tells us how much energy is required to liberate an electron from the metal surface. The rest of the energy goes into giving the electron kinetic energy.

Photoelectrons Ejection Rate

The rate at which photoelectrons are ejected is directly connected to the power of the light beam and the energy of the photons. The power of the light beam indicates how much energy is incident on the metal surface each second.
Given the power of the beam is 2.50 watts—or 2.50 joules per second—the number of photons, and thus photoelectrons, ejected per second can be calculated as follows:

• \( \text{Number of photoelectrons} = \frac{\text{Power}}{\text{Energy per photon}} \)

Using our earlier value for photon energy of \(1.601 \times 10^{-18}\) joules, we find around \(1.56 \times 10^{18}\) photoelectrons are ejected per second.
If either the power or wavelength of the light is altered, this rate changes. For example, halving the beam's power halves the photoelectron rate due to reduced photon incidence. Conversely, halving the wavelength while keeping power constant maintains the rate but alters the energy balance, thus potentially affecting whether electrons have enough energy to overcome the work function.