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The gap between valence and conduction bands in sillicon is 1.12 eV. A nickel nucleus in an excited state emits a gamma-ray photon with wavelength \(9.31 \times 10^{-4} \mathrm{nm}\) . How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

Short Answer

Expert verified
About \(1.20 \times 10^8\) electrons can be excited.

Step by step solution

01

Convert Wavelength to Energy

First, convert the wavelength of the gamma-ray photon into energy using the equation: \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant, \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 9.31 \times 10^{-4} \text{ nm} = 9.31 \times 10^{-13} \text{ m} \). Calculate to find \( E \).\[ E = \frac{(6.626 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{9.31 \times 10^{-13} \text{ m}} = 2.14 \times 10^{-11} \text{ J} \]
02

Convert Photon Energy to Electron Volts

Now, convert the energy from joules to electron volts since the gap is given in electron volts. Use the conversion factor: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \).\[ E = \frac{2.14 \times 10^{-11} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 1.34 \times 10^{8} \text{ eV} \]
03

Calculate Number of Excited Electrons

Finally, use the energy calculated in electron volts to determine how many electrons can be excited by one gamma-ray photon. Divide the photon's energy by the silicon band gap energy.\[ \text{Number of electrons} = \frac{1.34 \times 10^{8} \text{ eV}}{1.12 \text{ eV}} \approx 1.20 \times 10^{8} \]
04

Conclusion

The gamma-ray photon has enough energy to raise approximately \( 1.20 \times 10^8 \) electrons from the top of the valence band to the bottom of the conduction band in silicon.

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Key Concepts

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

Band Gap Energy
In solid-state physics, the band gap energy is a critical concept that pertains to the energy difference between the valence band and the conduction band of a material. The valence band is the highest range of electron energies in which electrons are normally present at absolute zero temperature, while the conduction band is the range where electron mobility increases, allowing for electrical conduction.
A material's band gap energy is a key factor in determining its electrical properties, such as whether it behaves as a conductor, insulator, or semiconductor. In semiconductors like silicon, this energy gap is relatively small, making it possible for electrons to jump from the valence band to the conduction band with modest energy input.
For silicon, the band gap energy is approximately 1.12 electron volts (eV). This value indicates how much energy is required to excite an electron from the valence band into the conduction band, enabling it to participate in conduction. This property is crucial in various technological applications, such as in the fabrication of electronic devices and solar cells.
Photon Energy
Photon energy is the amount of energy carried by a single photon, with its value depending upon the photon's wavelength or frequency. The relationship between photon energy and wavelength is inversely proportional—shorter wavelengths result in higher photon energies.
A commonly used formula to calculate the energy of a photon is \[ E = \frac{hc}{\lambda}, \]where:
  • \( E \) is the photon energy,
  • \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\),
  • \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\),
  • \( \lambda \) is the wavelength of the photon.
In our example, the gamma-ray photon has a wavelength of \(9.31 \times 10^{-4} \) nm, thus indicating a very high energy. Calculations reveal its energy to be approximately \(2.14 \times 10^{-11} \) J, which converts to about \(1.34 \times 10^8\) eV. This energy far exceeds the band gap of silicon, suggesting that a single gamma-ray photon can excite multiple electrons from the valence to conduction band.
Electron Excitation
Electron excitation refers to the process where an electron absorbs energy and moves to a higher energy state within an atom or a crystalline solid. In the context of band theory, this means an electron transitions from the valence band to the conduction band.
This process is pivotal in the functioning of semiconductors, where the excitation of electrons changes the electronic properties of the material, enabling devices to conduct electricity when needed. The energy required for electron excitation must at least equal the band gap energy of the material. For silicon, this is 1.12 eV.
When a gamma-ray photon with sufficiently high energy interacts with a material, it can transfer its energy to electrons. In silicon, the energy from such a photon, like the one discussed, is significant enough to excite approximately \(1.20 \times 10^8\) electrons all at once. This kind of electron movement is fundamental to various electronic and optoelectronic devices, including transistors, diodes, and light-emitting devices.

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Most popular questions from this chapter

If a sodium chloride (NaC) molecule could undergo an \(n \rightarrow n-1\) vibrational transition with no change in rotational quantum number, a photon with wavelength 20.0\(\mu \mathrm{m}\) would be emitted. The mass of a sodium atom is \(3.82 \times 10^{-26} \mathrm{kg}\) , and the mass of a chlorine atom is \(5.81 \times 10^{-26} \mathrm{kg}\) . Calculate the force constant \(k^{\prime}\) for the interatomic force in \(\mathrm{NaCl}\) .

Silver has a Fermi energy of 5.48 eV. Calculate the electron contribution to the molar heat capacity at constant volume of silver, \(C_{V},\) at 300 \(\mathrm{K}\) . Express your result \((\mathrm{a})\) as a multiple of \(R\) and (b) as a fraction of the actual value for silver, \(C_{V}=25.3 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) . (c) Is the value of \(C_{V}\) due principally to the electrons? If not, to what is it due? (Hint: See Section \(18.4 . )\)

(a) The equilibrium separation of the two nuclei in an NaCl molecule is 0.24 \(\mathrm{nm}\) If the molecule is modeled as charges \(+e\) and \(-e\) separated by 0.24 \(\mathrm{nm}\) , what is the electric dipole moment of the molecule (see Section 21.7\() ?\) (b) The measured electric dipole moment of an NaCl molecule is \(3.0 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} .\) If this dipole moment arises from point charges \(+q\) and \(-q\) separated by 0.24 \(\mathrm{nm}\) , what is \(q ?(\mathrm{c})\) A definition of the fractional ionic character of the bond is \(q | e .\) If the sodium atom has charge \(+e\) and the chlorine atom has charge \(-e\) , the fractional ionic character would be equal to \(1 .\) What is the actual fractional ionic character for the bond in NaCl?(d) The equilibrium distance between nuclei in the hydrogen iodide (HI) molecule is 0.16 \(\mathrm{nm}\) , and the measured electric dipole moment of the molecule is \(1.5 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) What is the fractional ionic character for the bond in HI? How does your answer compare to that for NaCl calculated in part (c)? Discuss reasons for the difference in these results.

The water molecule has an \(l=1\) rotational level \(1.01 \times 10^{-5} \mathrm{eV}\) above the \(l=0\) ground level. Calculate the wave length and frequency of the photon absorbed by water when it undergoes a rotational-level transition from \(l=0\) to \(l=1\) . The magnetron oscillator in a microwave oven generates microwaves with a frequency of 2450 \(\mathrm{MHz}\) Does this make sense, in view of the frequency you calculated in this problem? Explain.

Calculate the wavelengths of (a) a \(6.20-\mathrm{keV} \times\) ray; (b) a 37.6 -eV electron; (c) a 0.0205 -eV neutron.

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