# 7.2 The heisenberg uncertainty principle  (Page 3/6)

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## Significance

Based on early estimates of the size of a hydrogen atom and the uncertainty principle, the ground-state energy of a hydrogen atom is in the eV range. The ionization energy of an electron in the ground-state energy is approximately 10 eV, so this prediction is roughly confirmed. ( Note: The product $\hslash c$ is often a useful value in performing calculations in quantum mechanics.)

## Energy and time

Another kind of uncertainty principle concerns uncertainties in simultaneous measurements of the energy of a quantum state and its lifetime,

$\text{Δ}E\text{Δ}t\ge \frac{\hslash }{2},$

where $\text{Δ}E$ is the uncertainty in the energy measurement and $\text{Δ}t$ is the uncertainty in the lifetime measurement. The energy-time uncertainty principle    does not result from a relation of the type expressed by [link] for technical reasons beyond this discussion. Nevertheless, the general meaning of the energy-time principle is that a quantum state that exists for only a short time cannot have a definite energy. The reason is that the frequency of a state is inversely proportional to time and the frequency connects with the energy of the state, so to measure the energy with good precision, the state must be observed for many cycles.

To illustrate, consider the excited states of an atom. The finite lifetimes of these states can be deduced from the shapes of spectral lines observed in atomic emission spectra. Each time an excited state decays, the emitted energy is slightly different and, therefore, the emission line is characterized by a distribution of spectral frequencies (or wavelengths) of the emitted photons. As a result, all spectral lines are characterized by spectral widths. The average energy of the emitted photon corresponds to the theoretical energy of the excited state and gives the spectral location of the peak of the emission line. Short-lived states have broad spectral widths and long-lived states have narrow spectral widths.

## Atomic transitions

An atom typically exists in an excited state for about $\text{Δ}t={10}^{-8}\text{s}$ . Estimate the uncertainty $\text{Δ}\text{ }f$ in the frequency of emitted photons when an atom makes a transition from an excited state with the simultaneous emission of a photon with an average frequency of $f=7.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}\text{Hz}$ . Is the emitted radiation monochromatic?

## Strategy

We invert [link] to obtain the energy uncertainty $\text{Δ}E\approx \hslash \text{/}2\text{Δ}t$ and combine it with the photon energy $E=h\text{ }f$ to obtain $\text{Δ}\text{ }f$ . To estimate whether or not the emission is monochromatic, we evaluate $\text{Δ}f\text{/}f$ .

## Solution

The spread in photon energies is $\text{Δ}\text{ }E=h\text{Δ}\text{ }f$ . Therefore,

$\begin{array}{ccc}\hfill \text{Δ}E& \approx \hfill & \frac{\hslash }{2\text{Δ}t}⇒h\text{Δ}f\approx \frac{\hslash }{2\text{Δ}t}⇒\text{Δ}f\approx \frac{1}{4\text{π}\text{Δ}t}=\frac{1}{4\text{π}\left({10}^{-8}\text{s}\right)}=8.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{Hz}\text{,}\hfill \\ \hfill \frac{\text{Δ}f}{f}& =\hfill & \frac{8.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{Hz}}{7.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}\text{Hz}}=1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}.\hfill \end{array}$

## Significance

Because the emitted photons have their frequencies within $1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$ percent of the average frequency, the emitted radiation can be considered monochromatic.

Check Your Understanding A sodium atom makes a transition from the first excited state to the ground state, emitting a 589.0-nm photon with energy 2.105 eV. If the lifetime of this excited state is $1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\text{s}$ , what is the uncertainty in energy of this excited state? What is the width of the corresponding spectral line?

$4.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\text{eV}$ ; $1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{nm}$

## Summary

• The Heisenberg uncertainty principle states that it is impossible to simultaneously measure the x -components of position and of momentum of a particle with an arbitrarily high precision. The product of experimental uncertainties is always larger than or equal to $\hslash \text{/}2.$
• The limitations of this principle have nothing to do with the quality of the experimental apparatus but originate in the wave-like nature of matter.
• The energy-time uncertainty principle expresses the experimental observation that a quantum state that exists only for a short time cannot have a definite energy.

## Conceptual questions

If the formalism of quantum mechanics is ‘more exact’ than that of classical mechanics, why don’t we use quantum mechanics to describe the motion of a leaping frog? Explain.

Can the de Broglie wavelength of a particle be known precisely? Can the position of a particle be known precisely?

Yes, if its position is completely unknown. Yes, if its momentum is completely unknown.

Can we measure the energy of a free localized particle with complete precision?

Can we measure both the position and momentum of a particle with complete precision?

No. According to the uncertainty principle, if the uncertainty on the particle’s position is small, the uncertainty on its momentum is large. Similarly, if the uncertainty on the particle’s position is large, the uncertainty on its momentum is small.

## Problems

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

A gas of helium atoms at 273 K is in a cubical container with 25.0 cm on a side. (a) What is the minimum uncertainty in momentum components of helium atoms? (b) What is the minimum uncertainty in velocity components? (c) Find the ratio of the uncertainties in (b) to the mean speed of an atom in each direction.

a. $\text{Δ}p\ge 2.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{N}·\text{s}$ ; b. $\text{Δ}v\ge 6.31\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\text{m}$ ; c. $\text{Δ}v\text{/}\sqrt{{k}_{\text{B}}T\text{/}{m}_{\alpha }}=5.94\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}$

If the uncertainty in the $y$ -component of a proton’s position is 2.0 pm, find the minimum uncertainty in the simultaneous measurement of the proton’s $y$ -component of velocity. What is the minimum uncertainty in the simultaneous measurement of the proton’s $x$ -component of velocity?

Some unstable elementary particle has a rest energy of 80.41 GeV and an uncertainty in rest energy of 2.06 GeV. Estimate the lifetime of this particle.

$\text{Δ}\tau \ge 1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-25}\text{s}$

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.

Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1-eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon’s average frequency is this?

a. $\text{Δ}f\ge 1.59\phantom{\rule{0.2em}{0ex}}\text{MHz}$ ; b. $\text{Δ}\omega \text{/}{\omega }_{0}=3.135\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$

Suppose an electron is confined to a region of length 0.1 nm (of the order of the size of a hydrogen atom) and its kinetic energy is equal to the ground state energy of the hydrogen atom in Bohr’s model (13.6 eV). (a) What is the minimum uncertainty of its momentum? What fraction of its momentum is it? (b) What would the uncertainty in kinetic energy of this electron be if its momentum were equal to your answer in part (a)? What fraction of its kinetic energy is it?

as a free falling object increases speed what is happening to the acceleration
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
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As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
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law of multiple
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if we heated the ice then the refractive index be change from natural water
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Swati
yes
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The photoelectric effect is the emission of electrons when light shines on a material.