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Chapter 6: Problem 34

Titanium metal requires light with a maximum wavelength of \(286 \mathrm{nm}\) to emit electrons. (a) What is the minimum energy of the photons necessary to emit electrons from titanium via the photoelectric effect? (b) What is the frequency of this radiation? (c) Is it possible to eject electrons from titanium metal using infrared light? (d) If titanium is irradiated with light of wavelength \(276 \mathrm{nm}\), what is the maximum possible kinetic energy of the emitted electrons?

Short Answer

Expert verified
(a) 6.94 脳 10^{-19} J; (b) 1.05 脳 10^{15} Hz; (c) No, infrared can't eject electrons; (d) 0.26 脳 10^{-19} J.

Step by step solution

01

Convert Wavelength to Energy for Photons

To find the minimum energy of the photons necessary to emit electrons, we use the equation for energy derived from wavelength, which is given by \[ E = \frac{hc}{\lambda} \]where:- \( E \) is the energy of the photons,- \( h = 6.626 \times 10^{-34} \) J\cdot s is Planck's constant,- \( c = 3 \times 10^{8} \) m/s is the speed of light,- \( \lambda = 286 \times 10^{-9} \) m is the wavelength.Plug these values into the formula: \[ E = \frac{(6.626 \times 10^{-34} \, \text{J\cdot s})(3 \times 10^{8} \, \text{m/s})}{286 \times 10^{-9} \, \text{m}} \approx 6.94 \times 10^{-19} \text{ J} \].
02

Convert Wavelength to Frequency

The frequency of radiation, \( u \), can be found using the speed of light equation \( c = \lambda u \). Solve for frequency: \[ u = \frac{c}{\lambda} \]Substitute \( c = 3 \times 10^{8} \) m/s and \( \lambda = 286 \times 10^{-9} \) m:\[ u = \frac{3 \times 10^{8} \, \text{m/s}}{286 \times 10^{-9} \, \text{m}} \approx 1.05 \times 10^{15} \text{ Hz} \].
03

Check Possibility with Infrared Light

Infrared light has a longer wavelength than 286 nm. Longer wavelengths correspond to lower frequencies and thus lower energy. Therefore, infrared light, which has a wavelength much greater than 286 nm, cannot provide the minimum energy of \( 6.94 \times 10^{-19} \) J necessary to emit electrons from titanium.
04

Calculate Maximum Kinetic Energy of Electrons

If titanium is irradiated with light of wavelength 276 nm, calculate the energy of the photons using \[ E_{photon} = \frac{hc}{\lambda} \].For \( \lambda = 276 \times 10^{-9} \) m:\[ E_{photon} = \frac{(6.626 \times 10^{-34} \text{ J\cdot s})(3 \times 10^{8} \text{ m/s})}{276 \times 10^{-9} \text{ m}} \approx 7.20 \times 10^{-19} \text{ J} \].The work function \( \phi \) or threshold energy is the same as the minimum energy calculated in Step 1, \( 6.94 \times 10^{-19} \text{ J} \). The maximum kinetic energy \( KE_{max} \) can be calculated from \[ KE_{max} = E_{photon} - \phi \].\[ KE_{max} = 7.20 \times 10^{-19} \text{ J} - 6.94 \times 10^{-19} \text{ J} = 0.26 \times 10^{-19} \text{ J} \].

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Key Concepts

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

Photon Energy
The photoelectric effect describes the release of electrons from a metal surface when it is exposed to light or other electromagnetic radiation. This phenomenon illustrates the particle nature of light, where light is composed of packets of energy called photons. The energy of these photons is a crucial factor in determining their ability to eject electrons from a metal surface. This energy, known as "photon energy," is determined by the equation
\[ E = \frac{hc}{\lambda} \]
- Where: - \( E \) is the photon energy in joules (J), - \( h = 6.626 \times 10^{-34} \) J路s is Planck's constant, - \( c = 3 \times 10^{8} \) m/s is the speed of light, - \( \lambda \) is the wavelength.When calculating the photon energy, it's essential to convert the wavelength into meters by multiplying the given wavelength in nanometers by \( 10^{-9} \). Once you have the correct values, you can simply plug them into the formula. For metals like titanium, there is a specific minimum energy that photons must have to release electrons, known as the work function. If the photon energy exceeds this work function, electrons are emitted.
Kinetic Energy of Electrons
Once electrons are emitted from a metal due to the photoelectric effect, they can have kinetic energy. The kinetic energy of the emitted electrons depends on two main factors: the energy of the incoming photons and the work function of the metal. The work function is the minimum amount of energy required to release an electron from the metal's surface.The kinetic energy \( KE \) of the electrons can be calculated using the formula:
\[ KE_{max} = E_{photon} - \phi \]
- Where: - \( KE_{max} \) is the maximum kinetic energy of the emitted electrons, - \( E_{photon} \) is the energy of the incident photons, - \( \phi \) is the work function or threshold energy.This equation shows that the kinetic energy of electrons is the result of the excess energy the photon has beyond what is needed to overcome the work function. If the photon's energy is exactly equal to the work function, the electrons will have zero kinetic energy, meaning they are ejected but without any speed. If the photon energy is higher, the additional energy translates into the kinetic energy of the electrons, allowing them to move faster.
Wavelength and Frequency Relation
The relationship between a photon's wavelength, frequency, and energy is fundamental to understanding the photoelectric effect. Wavelength and frequency are inversely proportional, meaning that as one increases, the other decreases. This relationship is described by the equation:
\[ c = \lambda u \]
- Where: - \( c = 3 \times 10^{8} \) m/s is the speed of light, - \( \lambda \) is the wavelength of the light in meters, - \( u \) is the frequency in hertz (Hz).To find the frequency of a given wavelength, you can rearrange the formula to:
\[ u = \frac{c}{\lambda} \]
Substituting the values from the titanium metal problem, you can determine the frequency of radiation that corresponds to specific wavelengths. This is essential, as the frequency directly relates to the photon energy.
Since photon energy and frequency are directly proportional (as shown in \( E = hu \)), knowing one allows you to calculate the other. Thus, higher frequency light has more energetic photons and can cause ejection of electrons more effectively in comparison to lower frequency (or longer wavelength) light such as infrared, which, as established, does not possess sufficient energy to affect titanium.

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Most popular questions from this chapter

(a) For an \(\mathrm{He}^{+}\) ion, do the \(2 s\) and \(2 p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?

The first 25 years of the twentieth century were momentous for the rapid pace of change in scientists' understanding of the nature of matter. (a) How did Rutherford's experiments on the scattering of \(\alpha\) particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals preceding Thomson's work that the cathode rays are a wave phenomenon?

The following do not represent valid ground-state electron configurations for an atom either because they violate the Pauli exclusion principle or because orbitals are not filled in order of increasing energy. Indicate which of these two principles is violated in each example. (a) \([\mathrm{Ne}] 3 s^{2} 3 p^{6} 3 d^{5}\) (c) \(1 s^{2} 3 s^{1}\) (b) \([\mathrm{Xe}] 6 s^{3}\)

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm} .\) What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Determine which of the following statements are false and correct them. (a) The frequency of radiation increases as the wavelength increases. (b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of wavelength. (c) Infrared light has higher frequencies than visible light. (d) The glow from a fireplace, the energy within a microwave oven, and a foghorn blast are all forms of electromagnetic radiation.
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