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Chapter 28: Problem 20

A hydrogen atom has a radius of about \(0.05 \mathrm{nm}\) (a) Estimate the uncertainty in any component of the momentum of an electron confined to a region of this size. (b) From your answer to (a), estimate the electron's kinetic energy. (c) Does the estimate have the correct order of magnitude? (The ground-state kinetic energy predicted by the Bohr model is \(13.6 \mathrm{eV} .\) )

Short Answer

Expert verified
Answer: The estimated uncertainty in any component of the momentum of an electron confined to a hydrogen atom with a radius of 0.05 nm is approximately 1.057 x 10^-24 kg⋅m/s.

Step by step solution

01

Apply Heisenberg's Uncertainty Principle to find the uncertainty in momentum.

Heisenberg's Uncertainty Principle states that: \(∆x * ∆p ≥ \frac{h}{4π}\) Where ∆x is the uncertainty in position, ∆p is the uncertainty in momentum, and h is Planck's constant (approximately \(6.626 * 10^{-34} \mathrm{J\cdot s}\)). Given the radius of the hydrogen atom (0.05 nm), we can assume that the uncertainty in the position of the electron is roughly the same as the atom's radius: \(∆x = 0.05 \times 10^{-9} \mathrm{m}\) Now, solve for the uncertainty in momentum, ∆p: \(∆p ≥ \frac{h}{4π ∆x}\)
02

Calculate the uncertainty in momentum.

Plug in the values for h and ∆x: \(∆p ≥ \frac{6.626 \times 10^{-34} \mathrm{J\cdot s}}{4π (0.05 \times 10^{-9} \mathrm{m})}\) Upon calculation, we obtain: \(∆p ≥ 1.057 \times 10^{-24} \mathrm{kg\cdot m/s}\)
03

Estimate the electron's kinetic energy.

To estimate the electron's kinetic energy (KE), we can use the classical equation: \(KE = \frac{1}{2}mv^2\) where m is the mass of the electron and v is its velocity. We can rewrite this equation in terms of momentum (∆p) as follows: \(KE = \frac{∆p^2}{2m}\) The mass of an electron (m) is approximately \(9.11 \times 10^{-31} \mathrm{kg}\). Substitute the uncertainty in momentum, and mass of the electron: \(KE = \frac{(1.057 \times 10^{-24} \mathrm{kg\cdot m/s})^2}{2(9.11 \times 10^{-31} \mathrm{kg})}\) Upon calculation, we obtain: \(KE = 6.144 \times 10^{-19} \mathrm{J}\) Converting the kinetic energy to electron volts (eV), we get: \(KE = 3.84 \mathrm{eV}\)
04

Compare the estimate to the Bohr model's prediction.

We have estimated the electron's kinetic energy as 3.84 eV, which is on the order of magnitude of 1 eV. The Bohr model predicts the ground-state kinetic energy to be 13.6 eV which is on the order of magnitude of 10 eV. Since the two values differ by only one order of magnitude, our estimate can be considered to have the correct order of magnitude.

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Most popular questions from this chapter

A beam of neutrons is used to study molecular structure through a series of diffraction experiments. A beam of neutrons with a wide range of de Broglie wavelengths comes from the core of a nuclear reactor. In a time-offlight technique, used to select neutrons with a small range of de Broglie wavelengths, a pulse of neutrons is allowed to escape from the reactor by opening a shutter very briefly. At a distance of \(16.4 \mathrm{m}\) downstream, a second shutter is opened very briefly 13.0 ms after the first shutter. (a) What is the speed of the neutrons selected? (b) What is the de Broglie wavelength of the neutrons? (c) If each shutter is open for 0.45 ms, estimate the range of de Broglie wavelengths selected.
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A beam of neutrons has the same de Broglie wavelength as a beam of photons. Is it possible that the energy of each photon is equal to the kinetic energy of each neutron? If so, at what de Broglie wavelength(s) does this occur? [Hint: For the neutron, use the relativistic energy-momentum relation \(\left.E^{2}=E_{0}^{2}+(p c)^{2} .\right]\)
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