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schepotkina [342]
3 years ago
6

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

hography and the resulting ability to print ever-smaller features. Modern circuits are made using a variety of complicated lithography techniques, with the goal to make electronic traces as small and as close to each other as possible (to reduce the overall size, and thus increase the speed). In the end, though, all-optical techniques are limited by diffraction.
Assume we have a scannable laser that draws a line on a circuit board (the light exposes a line of photoresist, which then becomes impervious to a subsequent chemical etch, leaving only the narrow metal line under the exposed photoresist). Assume the laser wavelength is 248.0 nm (Krypton Fluoride excimer laser), the initial beam diameter is 1.0 cm, and the focusing lens (diameter = 1.3 cm) is extremely 'fast', with a focal length of only 0.625 cm.
a. What is the approximate width w of the line?
b. What is the minimum resolvable line separation between adjacent lines?
c. If the laser wavelength is instead reduced to 157 nm, what is the new minimum resolvable line separation?
Physics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

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A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

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4 0
3 years ago
The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl
Aleks04 [339]

The capacitance of the capacitor is 1.8\cdot 10^{-9}F

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the medium

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

d=3.5\cdot 10^{-5}m is the separation between the plates

A=0.063m \cdot 0.054m = 0.0034 m^2 (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the  capacitor is

C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F

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3 years ago
What is conductivity?​
kirill [66]

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Conductivity is the measure of the ease at which an electric charge or heat can pass through a material. A conductor is a material that gives very little resistance to the flow of an electric current or thermal energy.

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Immediately after it has started to rain, dust particles and oil residue float on the water and ____________ the traction of tir
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When it rains, dust particles and oil residues float on the water and this reduces the traction of tires.

<h3>What is traction?</h3>

This concept refers to a force between the tires and road that causes the movement of the wheels or vehicle is slower.

<h3>What happens with traction when it rains?</h3>

It is well-known more accidents occur when it rains, which is caused by cars slidding on the road. This is because when it rains traction or the grip of the wheel drastically reduces.

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Why do we always use water as a reference material when determining the relative density of a substance
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Answer:

Specific gravity for liquids is nearly always measured with respect to water at its densest

Explanation:

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