Problem 50 - College Physics

Question

A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects an electron. (a) What is the work function (in electronvolts) 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) 5.84 eV; (b) \(1.56 \times 10^{18}\); (c) \(7.8 \times 10^{17}\); (d) \(7.81 \times 10^{17}\).

Step 1: Calculate Energy of a Photon

The energy of a photon is given by the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.63 \times 10^{-34} \mathrm{~J \cdot s} \), \( c \) is the speed of light \( 3 \times 10^8 \mathrm{~m/s} \), and \( \lambda = 124 \mathrm{~nm} = 124 \times 10^{-9} \mathrm{~m} \). Calculate the energy \( E \):\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{124 \times 10^{-9}} \mathrm{~J} \approx 1.60 \times 10^{-18} \mathrm{~J} \]Convert this energy to electronvolts \((1 \mathrm{~eV} = 1.6 \times 10^{-19} \mathrm{~J})\):\[ E \approx \frac{1.60 \times 10^{-18}}{1.6 \times 10^{-19}} = 10 \mathrm{~eV} \]

Step 2: Calculate the Work Function

The work function \( \phi \) is the energy needed to release an electron from the metal and is related to the photon's energy and the maximum kinetic energy \( K_{\text{max}} \) of the ejected electrons by \( E = \phi + K_{\text{max}} \). Solve for the work function:\[ \phi = E - K_{\text{max}} = 10 \mathrm{~eV} - 4.16 \mathrm{~eV} = 5.84 \mathrm{~eV} \]

Step 3: Calculate Number of Photoelectrons Ejected per Second

The power of the beam \( P = 2.50 \mathrm{~W} \). Since each photon ejects one electron, the number of photoelectrons \( n \) per second is given by the formula \( n = \frac{P}{E} \). Convert the energy of each photon back to joules:\[ E = 1.60 \times 10^{-18} \mathrm{~J} \]Then calculate \( n \):\[ n = \frac{2.50}{1.60 \times 10^{-18}} \approx 1.56 \times 10^{18} \] photoelectrons per second.

Step 4: Effect of Reducing Power by Half

If the power is reduced by half, the new power is \( P' = 1.25 \mathrm{~W} \). The energy of a photon remains the same, so the number of photoelectrons ejected per second becomes:\[ n = \frac{1.25}{1.60 \times 10^{-18}} \approx 0.78 \times 10^{18} = 7.8 \times 10^{17} \]Therefore, if the power is reduced by half, the number of photoelectrons ejected each second is approximately \( 7.8 \times 10^{17} \).

Step 5: Effect of Reducing Wavelength by Half

Reducing the wavelength by half changes the photon energy. New wavelength \( \lambda' = 62 \mathrm{~nm} \), so new photon energy:\[ E' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{62 \times 10^{-9}} \approx 3.20 \times 10^{-18} \mathrm{~J} = 20 \mathrm{~eV} \]The number of photoelectrons ejected per second is:\[ n = \frac{2.50}{3.20 \times 10^{-18}} \approx 7.81 \times 10^{17} \]If the wavelength is reduced by half, approximately \( 7.81 \times 10^{17} \) photoelectrons are ejected each second.

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon. It is directly proportional to the frequency of the photon and inversely proportional to its wavelength. This can be defined mathematically by the formula \( E = \frac{hc}{\lambda} \), where:

• \( E \) is the energy of the photon
• \( h \) is Planck's constant, a fundamental constant of nature
• \( c \) is the speed of light
• \( \lambda \) is the wavelength of the light

The energy of a photon explains why light can exert force or energy on objects, even though individual photons are very small. When light of a specific wavelength hits a surface, its photons transfer energy to the electrons on that surface.
If the energy transferred is sufficient, it can eject electrons from the surface, leading to the photoelectric effect. The energy of the photons in this scenario can be measured in electronvolts (eV), which is a convenient unit of energy when dealing with atomic and subatomic particles.

Work Function

The work function of a material is the minimum energy required to remove an electron from the surface of a material, often a metal. In the context of the photoelectric effect, the work function is a key component. The photoelectric effect occurs when photons hit a material and transfer their energy to electrons, potentially freeing them if their energy meets or exceeds the work function.
The equation for the photoelectric effect is given by:

• \( E = \phi + K_{\text{max}} \)
• where \( E \) is the photon energy
• \( \phi \) is the work function
• \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electrons

This relationship helps in determining the work function given the energy of incoming photons and the kinetic energy of the ejected electrons. If the photon's energy is just equal to the work function, electrons are freed with no kinetic energy.
However, with higher photon energies, the excess energy is converted into kinetic energy for the ejected electrons.

Planck's Constant

Planck's constant \( h \) is a fundamental physical constant. It plays a vital role in the realm of quantum mechanics and is central to the description of the photoelectric effect. Its value is \( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \), which might seem small, but it is crucial for calculations involving quantum phenomena.
Planck's constant relates the energy of a photon to its frequency, showing that electromagnetic energy is quantized—a core principle of quantum theory. It appears in the equation for photon energy: \( E = h u \), where \( u \) is the frequency of the photon.

• With high frequency (small wavelength) light, Planck's constant ensures detectable levels of energy per photon.
• It's central to understanding why low-frequency light cannot eject electrons from certain materials, regardless of intensity.

This constant's importance goes beyond light and encompasses various aspects of quantum physics, governing processes on the atomic scale and ensuring stability in the universe's smallest constituents.