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AveGali [126]
1 year ago
8

transmission electron microscopes that use high-energy electrons accelerated over a range from 40.0 to 100 kv are employed in ma

ny applications including the study of biological samples (like a virus) and nanoscience research and development (alloy particles and carbon nanotubes, for example). what would be the spatial limitation (in pm) for this range of electrons? it is often true that resolution is limited by the optics of the lens system, not by the intrinsic limitation due to the de broglie wavelength.
Physics
1 answer:
Gekata [30.6K]1 year ago
7 0

The spatial limitations in Picometer for the given range of electrons would be around 50 picometers.

What is a transmission electron microscope?

A transmission electron microscope (TEM) is a type of microscope that uses a beam of high-energy electrons to produce detailed images of the structure of materials at the atomic or molecular scale. TEMs work by passing a focused beam of electrons through a thin sample and collecting the transmitted electrons on a fluorescent screen or an electronic detector. The interaction of the sample with the electrons results in the formation of an image that can be magnified and displayed on a computer monitor. TEMs are widely used in the fields of materials science, biology, and nanotechnology and can provide information about the structure, composition, and properties of materials with a high level and resolution.

According to the problem:

The spatial resolution of a transmission electron microscope (TEM) is determined by the size of the electron probe, which is directly related to the energy of the electrons. The higher the energy of the electrons is, the smaller the size of the probe is and the higher the spatial resolution.

At the lower end of the energy range of 40.0 kV, the spatial resolution of the TEM would be on the order of hundreds of nanometers. At the higher end of the range (100 kV), the spatial resolution would be on the order of tens of nanometers.

In general, TEMs with electron energy in the range of 40-100 kV are capable of resolving details down to around 50 picometers (pm). However, the actual spatial resolution will depend on various factors, such as the quality of the electron optics, the stability of the electron beam, and the sample preparation.

It's worth noting that TEMs with even higher electron energies (up to several hundred kV) are available, which can achieve spatial resolutions down to the sub-angstrom level (less than 0.1 pm). However, these instruments are much more expensive and complex to operate than TEMs with lower electron energies.

To know more about de broglie wavelength, visit:

brainly.com/question/17295250

#SPJ4

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A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
NeTakaya
Answer:
387 volts

Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)

In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm

Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts

Hope this helps :)
4 0
3 years ago
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Which of ONE of the following four elements has the most metallic properties?
olchik [2.2K]

Answer:

12 (Magnesium- Mg)

Explanation:

Looking at the four numbers, we have:

Magnesium, Silicon, Sulfur, and Chlorine.

We can eliminate two of the answers immediately just by looking at the periodic table.

Sulfur and Chlorine are on the nonmetal side of the periodic table. So that's <em>definitely</em> not it. That leaves Magnesium and Silicon.

Silicon is a Metalloid. Magnesium is an Alkaline earth Metal.

Metaloids are elements that have a mix of both<em> metal</em> and<em> nonmetal </em>properties (luster, how it feels, etc.). Since it's a MIX and Magnesium is just straight METAL-

We can say Magnesium has the most metallic properties.

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7 0
3 years ago
A student was traveling to see his grandmother who lives 15 miles north oh his home he started from rest and maintained a pace o
N76 [4]

Explanation:

Given parameters:

Distance  = 15miles north = 24140.2m

Initial velocity  = 0m/s

Final velocity  = 4m/s

Unknown:

Speed, velocity and acceleration = ?

Solution:

The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.

  Speed  = \frac{distance }{time}  

  The speed of the student is 4m/s

Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;

    Velocity  = \frac{displacement}{time}  

   The velocity of the student is 4m/s due north

Acceleration is the change in velocity with time;

     To find the acceleration, we use

      v²  = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

     4² = 0² + 2x a x 24140.2

       a  = \frac{16}{2 x 24140.2}  = 0.00033m/s²

3 0
2 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

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