# 6.6 Wave-particle duality  (Page 4/12)

 Page 4 / 12

Such limitations do not appear in the scanning electron microscope (SEM) , which was invented by Manfred von Ardenne in 1937. In an SEM, a typical energy of the electron beam is up to 40 keV and the beam is not transmitted through a sample but is scattered off its surface. Surface topography of the sample is reconstructed by analyzing back-scattered electrons, transmitted electrons, and the emitted radiation produced by electrons interacting with atoms in the sample. The resolving power of an SEM is better than 1 nm, and the magnification can be more than 250 times better than that obtained with a light microscope. The samples scanned by an SEM can be as large as several centimeters but they must be specially prepared, depending on electrical properties of the sample.

High magnifications of the TEM and SEM allow us to see individual molecules. High resolving powers of the TEM and SEM allow us to see fine details, such as those shown in the SEM micrograph of pollen at the beginning of this chapter ( [link] ).

## Resolving power of an electron microscope

If a 1.0-pm electron beam of a TEM passes through a $2.0\text{-}\mu \text{m}$ circular opening, what is the angle between the two just-resolvable point sources for this microscope?

## Solution

We can directly use a formula for the resolving power, $\text{Δ}\theta ,$ of a microscope (discussed in a previous chapter) when the wavelength of the incident radiation is $\lambda =1.0\phantom{\rule{0.2em}{0ex}}\text{pm}$ and the diameter of the aperture is $D=2.0\mu \text{m}:$

$\text{Δ}\theta =1.22\frac{\lambda }{D}=1.22\frac{1.0\phantom{\rule{0.2em}{0ex}}\text{pm}}{2.0\mu \text{m}}=6.10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\text{rad}=3.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{degree.}$

## Significance

Note that if we used a conventional microscope with a 400-nm light, the resolving power would be only $14\text{°},$ which means that all of the fine details in the image would be blurred.

Check Your Understanding Suppose that the diameter of the aperture in [link] is halved. How does it affect the resolving power?

doubles it

## Summary

• Wave-particle duality exists in nature: Under some experimental conditions, a particle acts as a particle; under other experimental conditions, a particle acts as a wave. Conversely, under some physical circumstances, electromagnetic radiation acts as a wave, and under other physical circumstances, radiation acts as a beam of photons.
• Modern-era double-slit experiments with electrons demonstrated conclusively that electron-diffraction images are formed because of the wave nature of electrons.
• The wave-particle dual nature of particles and of radiation has no classical explanation.
• Quantum theory takes the wave property to be the fundamental property of all particles. A particle is seen as a moving wave packet. The wave nature of particles imposes a limitation on the simultaneous measurement of the particle’s position and momentum. Heisenberg’s uncertainty principle sets the limits on precision in such simultaneous measurements.
• Wave-particle duality is exploited in many devices, such as charge-couple devices (used in digital cameras) or in the electron microscopy of the scanning electron microscope (SEM) and the transmission electron microscope (TEM).

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
how does colour appear in thin films
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
derivation of lateral shieft
hi
Imran
total binding energy of ionic crystal at equilibrium is
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Sushant
What is motion
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
how can u say that
Chemist
do u know about black hole
Chemist
Not so more
Sushant
DHEERAJ
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
acceleration also increases
Jay
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Michael
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
Big
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?