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Chapter 28: Problem 27

What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)

Short Answer

Expert verified
Answer: The minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.

Step by step solution

01

Calculate uncertainty in momentum (∆p)

Using Heisenberg Uncertainty Principle, ∆x * ∆p ≥ h/(4π) Given: ∆x = 1.0 fm = 1.0 × 10^(-15) m We want to find the minimum possible uncertainty in momentum (∆p), so we must find the equality. ∆p = (h/(4π))/ ∆x Now, substitute the known values and the given ∆x value: ∆p = ((6.626 × 10^(-34) Js)/(4π)) / (1.0 × 10^(-15) m)
02

Calculate the minimum kinetic energy (K.E.)

The formula for kinetic energy is: K.E. = p^2/(2m) The minimum kinetic energy will occur when the momentum (p) equals the minimum possible uncertainty in momentum (∆p). So, we can rewrite the formula as: K.E_min = (∆p)^2 / (2m) Substitute the value of ∆p from Step 1 and the mass of the electron (m = 9.11 × 10^(-31) kg): K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg))
03

Calculate the numerical value of the minimum kinetic energy

Now, we can plug in the numbers and calculate the minimum kinetic energy. K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg)) K.E_min ≈ 1.763 × 10^(-14) J Thus, the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.

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Most popular questions from this chapter

What is the ground-state electron configuration of tellurium (Te, atomic number 52 )?
A beam of electrons is accelerated across a potential of \(15 \mathrm{kV}\) before passing through two slits. The electrons form a interference pattern on a screen \(2.5 \mathrm{m}\) in front of the slits. The first-order maximum is \(8.3 \mathrm{mm}\) from the central maximum. What is the distance between the slits?
Before the discovery of the neutron, one theory of the nucleus proposed that the nucleus contains protons and electrons. For example, the helium-4 nucleus would contain 4 protons and 2 electrons instead of - as we now know to be true- 2 protons and 2 neutrons. (a) Assuming that the electron moves at nonrelativistic speeds, find the ground-state energy in mega-electron- volts of an electron confined to a one-dimensional box of length $5.0 \mathrm{fm}\( (the approximate diameter of the \)^{4} \mathrm{He}$ nucleus). (The electron actually does move at relativistic speeds. See Problem \(80 .)\) (b) What can you conclude about the electron-proton model of the nucleus? The binding energy of the \(^{4} \mathrm{He}\) nucleus - the energy that would have to be supplied to break the nucleus into its constituent particles-is about \(28 \mathrm{MeV} .\) (c) Repeat (a) for a neutron confined to the nucleus (instead of an electron). Compare your result with (a) and comment on the viability of the proton-neutron theory relative to the electron-proton theory.
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A light-emitting diode (LED) has the property that electrons can be excited into the conduction band by the electrical energy from a battery; a photon is emitted when the electron drops back to the valence band. (a) If the band gap for this diode is \(2.36 \mathrm{eV},\) what is the wavelength of the light emitted by the LED? (b) What color is the light emitted? (c) What is the minimum battery voltage required in the electrical circuit containing the diode?
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