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Chapter 7: Problem 32

A velocity measurement of an \(\alpha\) -particle has been performed with a precision of \(0.02 \mathrm{mm} / \mathrm{s}\). What is the minimum uncertainty in its position?

Short Answer

Expert verified
The minimum uncertainty in the position of the alpha particle is approximately \(2.635 \times 10^{-32} \, \text{m}\).

Step by step solution

01

Identify the precision in velocity measurement

The precision in the velocity measurement is given as 0.02 mm/s. Let's first convert this value from millimeters per second to meters per second. Precision in velocity measurement: \(0.02 \, \text{mm/s} = 0.02 \times 10^{-3} \, \text{m/s}\)
02

Write down Heisenberg's Uncertainty Principle for position and velocity

Heisenberg's Uncertainty Principle states that the product of the uncertainty in position, \(\Delta x\), and the uncertainty in velocity, \(\Delta v\), is greater than or equal to the reduced Planck's constant divided by 2: \(\Delta x \cdot \Delta v \geq \frac{\hbar}{2}\) where \(\hbar = \frac{h}{2 \pi}\) and h is Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{J s}\)
03

Calculate the reduced Planck's constant

Let's calculate the value of the reduced Planck's constant using the given value for Planck's constant: \(\hbar = \frac{6.626 \times 10^{-34} \, \text{J s}}{2 \pi} = 1.054 \times 10^{-34} \, \text{J s}\)
04

Calculate the minimum uncertainty in position

Now we can calculate the minimum uncertainty in position using the given precision in velocity and Heisenberg's Uncertainty Principle: \(\Delta x \geq \frac{\hbar}{2 \Delta v} = \frac{1.054 \times 10^{-34} \, \text{J s}}{2 \cdot 0.02 \times 10^{-3} \, \text{m/s}}\) \(\Delta x \geq 2.635 \times 10^{-32} \, \text{m}\) The minimum uncertainty in the position of the alpha particle is approximately \(2.635 \times 10^{-32} \, \text{m}\).

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Most popular questions from this chapter

Is it possible to measure energy of \(0.75 \hbar \omega\) for a quantum harmonic oscillator? Why? Why not? Explain.

. Estimate the ground state energy of the quantum harmonic oscillator by Heisenberg's uncertainty principle. Start by assuming that the product of the uncertainties \(\Delta x\) and \(\Delta p\) is at its minimum. Write \(\Delta p\) in terms of \(\Delta x\) and assume that for the ground state \(x \approx \Delta x\) and \(p \approx \Delta p\) then write the ground state energy in terms of \(x .\) Finally, find the value of \(x\) that minimizes the energy and find the minimum of the energy.

A diatomic molecule behaves like a quantum harmonic oscillator with the force constant \(12.0 \mathrm{N} / \mathrm{m}\) and mass \(5.60 \times 10^{-26} \mathrm{kg}\). (a) What is the wavelength of the emitted photon when the molecule makes the transition from the third excited state to the second excited state? (b) Find the ground state energy of vibrations for this diatomic molecule.

An atom in a metastable state has a lifetime of 5.2 ms. Find the minimum uncertainty in the measurement of energy of the excited state.

An electron is confined to a box of width \(0.25 \mathrm{nm}\). (a) Draw an energy-level diagram representing the first five states of the electron. (b) Calculate the wavelengths of the emitted photons when the electron makes transitions between the fourth and the second excited states, between the second excited state and the ground state, and between the third and the second excited states.
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