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Chapter 28: Problem 13

In a photoelectric-effect experiment, it is found that no current flows unless the incident light has a wavelength shorter than \(289 \mathrm{nm}\). (a) What is the work function of the metal surface? (b) What stopping potential will be needed to halt the current if light of \(225 \mathrm{nm}\) falls on the surface?

Short Answer

Expert verified
(a) The work function is \( 6.88 \times 10^{-19} \text{ J} \). (b) The stopping potential is \( 1.225 \text{ V} \).

Step by step solution

01

Understanding the Photoelectric Effect

The photoelectric effect refers to the emission of electrons from a metal surface when light of a certain frequency (threshold frequency) is incident on it. The work function \( \phi \) of the metal is the minimum energy needed to release electrons from its surface.
02

Calculate the Work Function

The threshold wavelength \( \lambda_0 \) is given as 289 nm. The energy of the photon at this threshold is equal to the work function \( \phi \). Use the equation \( \phi = \frac{hc}{\lambda_0} \), where \( h = 6.63 \times 10^{-34} \text{ J s} \) is Planck's constant, and \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. Convert \( \lambda_0 \) to meters, \( 289 \text{ nm} = 289 \times 10^{-9} \text{ m} \). Substitute the values to find \( \phi \).
03

Calculate the Stopping Potential

Light of wavelength 225 nm is incident on the surface. First, calculate the energy of the incident photons \( E = \frac{hc}{\lambda} \), where \( \lambda = 225 \times 10^{-9} \text{ m} \). The kinetic energy of the emitted electrons is given by \( KE = E - \phi \). The stopping potential \( V \) is given by \( KE = eV \), where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)). Solve for \( V \).
04

Solve the Equations

Substitute the known values and calculate: \( \phi = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{289 \times 10^{-9}} = 6.88 \times 10^{-19} \text{ J} \). For light at 225 nm, \( E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{225 \times 10^{-9}} = 8.84 \times 10^{-19} \text{ J} \). \( KE = 8.84 \times 10^{-19} - 6.88 \times 10^{-19} = 1.96 \times 10^{-19} \text{ J} \). \( V = \frac{1.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.225 \text{ V} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Function
The work function is a fundamental concept in the photoelectric effect. It represents the minimum energy required to release an electron from the surface of a metal when light is shined upon it. Imagine it as the energy barrier that the electrons must overcome to escape from the metal's surface. It's specific to each material, depending on its electronic structure.

To calculate the work function (\( \phi \)), we use the equation:
  • \[ \phi = \frac{hc}{\lambda_0} \]
Here,
  • \( h = 6.63 \times 10^{-34} \text{ Js} \) is Planck's constant.
  • \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light.
  • \( \lambda_0 \) is the threshold wavelength, the longest wavelength of light that can cause electrons to be emitted from the surface.
If light has a wavelength longer than the threshold wavelength, it does not have enough energy to overcome the work function, and no electrons are emitted. It is critical for determining the conditions under which the photoelectric effect occurs.
Stopping Potential
The stopping potential is the potential difference needed to stop the current caused by electrons emitted from the metal surface in the photoelectric effect. Once electrons are ejected, they carry kinetic energy due to the energy absorbed from the light. The stopping potential is applied to completely halt these electrons, preventing them from causing a current.

To find the stopping potential (\( V \)), we use:
  • \[ KE = eV \]
where
  • \( KE \) is the kinetic energy of the electrons, calculated as \( E - \phi \), with \( E \) being the energy of the incident photons.
  • \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)).
This concept illustrates how we can measure the energy given to the electrons by the light. By adjusting the stopping potential, we can determine the maximum kinetic energy of these photoelectrons, giving insights into the energy conversion taking place at the atomic level.
Planck's Constant
Planck's constant is crucial in understanding quantum mechanics and is prominently featured in the photoelectric effect. It links the energy of photons (light particles) to their frequency (how quickly the wave oscillates), defining the quantization of energy levels.

The equation connecting energy \( E \) to frequency \( f \) is:
  • \[ E = hf \]
where
  • \( h = 6.63 \times 10^{-34} \text{ Js} \) is Planck's constant.
This tiny but fundamental constant transforms our understanding of energy from a continuous spectrum into discrete packets or quanta, which are necessary for explaining phenomena at the micro level like the photoelectric effect. Understanding it is essential for several areas in physics such as quantum mechanics, spectroscopy, and many others, underlying the complex nature of light and matter.

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Most popular questions from this chapter

Removing vascular lesions. A pulsed dye laser emits light of wavelength \(585 \mathrm{nm}\) in \(450 \mu \mathrm{s}\) pulses. Because this wavelength is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as port- wine-colored birthmarks. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \(\left(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, 2.256 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right) .\) Suppose that each pulse must remove \(2.0 \mu \mathrm{g}\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\). (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

Response of the eye. The human eye is most sensitive to green light of wavelength \(505 \mathrm{nm} .\) Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina. (a) What is the frequency of this photon? (b) How much energy (in joules and eV) does it deliver to the receptor cells? (c) To appreciate what a small amount of energy this is, calculate how fast a typical bacterium of mass \(9.5 \times 10^{-12} \mathrm{~g}\) would move if it had that much energy.

If a photon of wavelength \(0.04250 \mathrm{nm}\) strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon, (b) the wavelength of the scattered light, (c) the change in energy of the photon (is it a loss or a gain?), and (d) the energy gained by the electron.

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)?

A photon with wavelength \(\lambda\) collides with a free electron. The scattered photon has a wavelength of \(\lambda^{\prime}=2 \lambda .\) If the incident photon has a wavelength of \(\lambda=1.0 \times 10^{-12} \mathrm{~m},\) through what angle is it Compton scattered?
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