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Chapter 6: Problem 30

Sodium metal requires a photon with a minimum energy of \(4.41 \times 10^{-19} \mathrm{~J}\) to emit electrons. (a) What is the minimum frequency of light necessary to emitelectrons from sodium via the photoelectric effect? (b) What is the wavelength of this light? (c) If sodium is irradiated with light of \(439 \mathrm{~nm}\), what is the maximum possible kinetic energy of the emitted electrons? (d) What is the maximum number of electrons that can be freed by a burst of light whose total energy is \(1.00 \mu] ?\)

Short Answer

Expert verified
(a) The minimum frequency of light required to emit electrons from sodium is \(6.65 \times 10^{14} Hz\). (b) The wavelength of this light is \(451 nm\). (c) The maximum possible kinetic energy of the emitted electrons when sodium is irradiated with light of \(439 nm\) is \(1.2 \times 10^{-20} J\). (d) The maximum number of electrons that can be freed by a burst of light with a total energy of \(1.00 \mu J\) is approximately \(2.27 \times 10^{12}\) electrons.

Step by step solution

01

(a) Finding the minimum frequency of light

To find the minimum frequency of light necessary to emit electrons from sodium, we can use Planck's equation \(E = h \nu\), where E is the minimum energy given and h is Planck's constant (\(6.63 \times 10^{-34} Js\)). Solving for \(\nu\), the equation becomes: \[\nu = \frac{E}{h}\] Plug in the given values of E and h: \[\nu = \frac{4.41 \times 10^{-19} J}{6.63 \times 10^{-34} Js} = 6.65 \times 10^{14} Hz\]
02

(b) Finding the wavelength of the light

Now we want to find the wavelength of the light for which the minimum frequency was derived. We can do that using the speed of light equation: \(c = \lambda \nu\), where c is the speed of light (\(3 \times 10^8 m/s\)) and \(\nu\) is the frequency found in step (a). Solving for \(\lambda\), the equation becomes: \[\lambda = \frac{c}{\nu}\] Plug in the values of c and \(\nu\): \[\lambda = \frac{3 \times 10^8 m/s}{6.65 \times 10^{14} Hz} = 4.51 \times 10^{-7} m = 451 nm\]
03

(c) Finding the maximum possible kinetic energy of the emitted electrons

Given that sodium is irradiated with a light of 439 nm, we should first find the energy of this photon. Using the speed of light equation: \[\nu = \frac{c}{\lambda}\] Plug in the values of c and \(\lambda\): \[\nu = \frac{3 \times 10^8 m/s}{439 \times 10^{-9} m} = 6.83 \times 10^{14} Hz\] Now, we can find the energy of the photon by using Planck's equation: \[E = h \nu\] Plug in the values of h and \(\nu\): \[E = 6.63 \times 10^{-34} Js \times 6.83 \times 10^{14} Hz = 4.53 \times 10^{-19} J\] Now we can find the maximum possible kinetic energy (K.E.) by subtracting the minimum energy required to emit electrons from the energy of the given photon: \[K.E. = E - E_{min}\] Plug in the known energy values: \[K.E. = 4.53 \times 10^{-19} J - 4.41 \times 10^{-19} J = 1.2 \times 10^{-20} J\]
04

(d) Finding the maximum number of electrons freed by a burst of light

The total energy of the burst of light is given as \(1.00 \times 10^{-6} J\). Using the minimum energy required to emit one electron from the sodium metal, we can find the maximum number of electrons that can be freed. Let n be the number of electrons: \[E_{total} = n(E_{min})\] Solving for n: \[n = \frac{E_{total}}{E_{min}}\] Plug in the given values of the total energy and minimum energy: \[n = \frac{1.00 \times 10^{-6} J}{4.41 \times 10^{-19} J} \approx 2.27 \times 10^{12}\] The maximum number of electrons that can be freed by a burst of light is approximately \(2.27 \times 10^{12}\) electrons.

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Key Concepts

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

Planck's Constant
One of the fundamental constants in physics, Planck's constant, denoted by \( h \), plays a critical role in the study of quantum mechanics and the photoelectric effect. It's a tiny number, valued at approximately \( 6.63 \times 10^{-34} \text{ Js} \).
This constant is crucial because it relates the energy of a photon to its frequency. The photoelectric effect, which involves ejecting electrons from a material when light shines on it, relies heavily on Planck's constant.
  • Energy and frequency are connected by Planck鈥檚 equation: \( E = h u \).
  • In the equation, \( E \) is energy, \( u \) is frequency, and \( h \) is Planck's constant.
Understanding Planck's constant helps us determine how much energy each photon carries, which is necessary to overcome the work function of a metal and release electrons.
Energy of a Photon
Photons are the particles of light and each carries a quantifiable amount of energy. This energy is not dependent on the intensity of light, but rather on its frequency. This means high-frequency light waves, such as ultraviolet light, have more energetic photons than low-frequency waves like infrared.
The energy of a single photon \( E \) can be calculated using the formula \( E = h u \), which as mentioned before, links energy with frequency via Planck鈥檚 constant \( h \).
  • Photon energy determines whether a material's electrons can be ejected when exposed to light.
  • For sodium, as seen in the problem, the energy must be at least \( 4.41 \times 10^{-19} \text{ J} \) to emit electrons.
Realizing how photon energy is computed reveals why certain frequencies are needed to initiate the photoelectric effect.
Kinetic Energy of Electrons
When photons strike a metal surface during the photoelectric effect, they transfer energy to electrons, causing these electrons to be ejected from the material. The kinetic energy \( KE \) of these emitted electrons can be calculated using the equation \( KE = E - E_{\text{min}} \), where \( E \) is the photon energy and \( E_{\text{min}} \) is the minimum energy needed to remove an electron, known as the work function.
From the problem, once you know the photon's energy and the work function, finding the kinetic energy is straightforward.
  • Emission occurs if photon energy exceeds the work function.
  • Higher photon energy leads to greater kinetic energy of ejected electrons.
The difference in energies determines how much kinetic energy an electron will have, affecting its speed once it leaves the metal.
Wavelength and Frequency Relationship
Light possesses both wave-like and particle-like properties. One essential wave property is wavelength \( \lambda \), which is the distance between consecutive peaks of the wave. There's a direct relationship between wavelength and frequency \( u \), described by the formula \( c = \lambda u \), where \( c \) is the constant speed of light \( 3 \times 10^8 \text{ m/s} \).
  • A higher frequency corresponds to a shorter wavelength.
  • Lower frequency means a longer wavelength.
These characteristics are inversely proportional, so when calculating either frequency or wavelength, adjusting one affects the other.
This relationship is key when determining the wavelength of light emitted, as it provides insight into the light's energy potential in applications like the photoelectric effect.

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Give the values for \(n, l\), and \(m_{l}\) for (a) each orbital in the \(2 p\) subshell, (b) each orbital in the \(5 d\) subshell.

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The discovery of hafnium, element number 72, provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and \(G\). von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 \mathrm{Write}\) a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

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