Assertion (A) : E-beam lithography is used for drawing nanostructures. Reason (R) : E-beam lithography has high resolution than photo lithography. Select your answer using the codes given below :
Both (A) and (R) are true and (R) is the correct explanation of (A)
Both statements are true, and the second is precisely why the first is true.
Resolution is limited by the wavelength of whatever writes the pattern. The diffraction limit is
\(R=k_{1}\dfrac{\lambda}{NA}\)
so shortening λ is the only fundamental way to draw finer features. Optical lithography uses deep-ultraviolet light at 248 or 193 nm, which sets its unaided limit at roughly a hundred nanometres.
An electron beam changes the wavelength by orders of magnitude. By de Broglie,
\(\lambda=\dfrac{h}{\sqrt{2mqV}}\)
and at an accelerating voltage of 50 kV this gives about 0.005 nm — some forty thousand times shorter than 193 nm light. Diffraction ceases to matter altogether, and the achievable resolution instead becomes limited by practical effects: electron scattering in the resist, backscatter from the substrate (the proximity effect), and the finite size of the focused spot. Features of 10 nm and below are routine.
| Photolithography | E-beam lithography | |
|---|---|---|
| Wavelength | 193 nm (DUV) | ~0.005 nm at 50 kV |
| Resolution | ~100 nm unaided | < 10 nm |
| Method | Whole wafer through a mask, at once | Serial — one spot at a time |
| Throughput | High | Very low |
| Mask | Required | None — direct write |
So (R) explains (A) directly, and the code is 1.
The cost of that resolution is speed, and it is what decides where each technique is used. Photolithography exposes an entire die in one flash; e-beam scans the pattern point by point, so writing a full wafer can take hours. That makes e-beam impractical for volume production but ideal for the two jobs it dominates: research into nanostructures such as quantum dots, nanowires and single-electron devices, and mask making — the photomasks used by optical lithography are themselves written by electron beam.
The proximity effect deserves a note because it is the practical limit rather than the theoretical one: electrons scatter forward through the resist and backscatter from the substrate, exposing resist adjacent to the intended spot. Dense patterns therefore receive more dose than isolated ones, and the pattern data must be dose-corrected before writing.
Hence, both (A) and (R) are true and (R) is the correct explanation of (A).
To form the integrated circuits, VLSI technology uses
The application of SiO2 in VLSI technology is mainly
A. to use it as a 'Mask'.
B. to provide electrical isolation.
C. to provide surface activation.
D. to provide electrical conduction.
E. as a etching material.
Choose the most appropriate answer from the options given below :
Match List I with List II
| LIST I ( IC classification) | LIST II (Number of transistors used) |
| A. VLSI | I. 1000 \(\lt\) Number of transistors \(\lt\) 10,000 |
| B. MSI | II. Number of transistors \(\gt\) 10,000 |
| C. LSI | III. Number of transistors \(\lt\) 100 |
| D. SSI | IV. 100 \(\lt\) Number of transistors \(\lt\) 1,000 |
Choose the correct answer from the options given below:
In VLSI Design flow,
A. Algorithm describes the behaviour of target chip.
B. Architecture of processor is not mapped onto the chip surface by floor planning.
C. Behavioral domain defines FSMs.
D. Individual modules are implemented with leaf cells.
Choose the correct answer from the options given below:
The correct design flow of VLSI is :
A. Functional Design & verification
B. Circuit Design & verification
C. Logic Design & verification
D. Physical Design
E. Layout verification
Choose the correct answer from the options given below :
Microprocessor development had happened because of LSI. What is LSI ?
When exposed to light, the positive resists in developer solution responds by becoming :
To form the integrated circuits, VLSI technology uses
The application of SiO2 in VLSI technology is mainly
A. to use it as a 'Mask'.
B. to provide electrical isolation.
C. to provide surface activation.
D. to provide electrical conduction.
E. as a etching material.
Choose the most appropriate answer from the options given below :
Match List I with List II
| LIST I ( IC classification) | LIST II (Number of transistors used) |
| A. VLSI | I. 1000 \(\lt\) Number of transistors \(\lt\) 10,000 |
| B. MSI | II. Number of transistors \(\gt\) 10,000 |
| C. LSI | III. Number of transistors \(\lt\) 100 |
| D. SSI | IV. 100 \(\lt\) Number of transistors \(\lt\) 1,000 |
Choose the correct answer from the options given below:
In VLSI Design flow,
A. Algorithm describes the behaviour of target chip.
B. Architecture of processor is not mapped onto the chip surface by floor planning.
C. Behavioral domain defines FSMs.
D. Individual modules are implemented with leaf cells.
Choose the correct answer from the options given below: