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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 脳 10^(-14) J or 17.63 MeV.

Step by step solution

01

Calculate uncertainty in momentum (鈭唒)

Using Heisenberg Uncertainty Principle, 鈭唜 * 鈭唒 鈮� h/(4蟺) Given: 鈭唜 = 1.0 fm = 1.0 脳 10^(-15) m We want to find the minimum possible uncertainty in momentum (鈭唒), so we must find the equality. 鈭唒 = (h/(4蟺))/ 鈭唜 Now, substitute the known values and the given 鈭唜 value: 鈭唒 = ((6.626 脳 10^(-34) Js)/(4蟺)) / (1.0 脳 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 (鈭唒). So, we can rewrite the formula as: K.E_min = (鈭唒)^2 / (2m) Substitute the value of 鈭唒 from Step 1 and the mass of the electron (m = 9.11 脳 10^(-31) kg): K.E_min = ((6.626 脳 10^(-34) Js / (4蟺))/(1.0 脳 10^(-15) m))^2 / (2 * (9.11 脳 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 脳 10^(-34) Js / (4蟺))/(1.0 脳 10^(-15) m))^2 / (2 * (9.11 脳 10^(-31) kg)) K.E_min 鈮� 1.763 脳 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 脳 10^(-14) J or 17.63 MeV.

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Most popular questions from this chapter

(a) Make a qualitative sketch of the wave function for the \(n=5\) state of an electron in a finite box \([U(x)=0\) for \(00\) elsewherel. (b) If \(L=1.0 \mathrm{nm}\) and \(U_{0}=1.0\) keV, estimate the number of bound states that exist.
An electron in a one-dimensional box has ground-state energy 0.010 eV. (a) What is the length of the box? (b) Sketch the wave functions for the lowest three energy states of the electron. (c) What is the wavelength of the electron in its second excited state \((n=3) ?\) (d) The electron is in its ground state when it absorbs a photon of wavelength \(15.5 \mu \mathrm{m}\). Find the wavelengths of the photon(s) that could be emitted by the electron subsequently.
A free neutron (that is, a neutron on its own rather than in a nucleus) is not a stable particle. Its average lifetime is 15 min, after which it decays into a proton, an electron, and an antineutrino. Use the energy-time uncertainty principle \([\mathrm{Eq} .(28-3)]\) and the relationship between mass and rest energy to estimate the inherent uncertainty in the mass of a free neutron. Compare with the average neutron mass of \(1.67 \times 10^{-27} \mathrm{kg} .\) (Although the uncertainty in the neutron's mass is far too small to be measured, unstable particles with extremely short lifetimes have marked variation in their measured masses.)
A radar pulse has an average wavelength of \(1.0 \mathrm{cm}\) and lasts for \(0.10 \mu \mathrm{s}\). (a) What is the average energy of the photons? (b) Approximately what is the least possible uncertainty in the energy of the photons?
A fly with a mass of \(1.0 \times 10^{-4} \mathrm{kg}\) crawls across a table at a speed of \(2 \mathrm{mm} / \mathrm{s} .\) Compute the de Broglie wavelength of the fly and compare it with the size of a proton (about $\left.1 \mathrm{fm}, 1 \mathrm{fm}=10^{-15} \mathrm{m}\right)$.
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