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Microscope and Microscopy - Scanning Electron Microscope (SEM)-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Scanning Electron Microscope (SEM) uses a focused beam of high-energy electrons to generate a variety of signals at the surface of solid specimens, revealing information about morphology, chemical composition, and crystalline structure.

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Unlike a Light Microscope that uses photons, the SEM uses electrons with a much smaller wavelength, allowing for much higher resolution and magnification. The wavelength λ\lambda of an electron is inversely proportional to its momentum.

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The primary signals used for imaging are Secondary Electrons (SESE) and Backscattered Electrons (BSEBSE). SESE are typically used for showing morphology and topography, while BSEBSE show contrast in composition (atomic number contrast).

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Specimens must be conductive to prevent the accumulation of static electric charge. Non-conductive materials are usually coated with an ultra-thin layer of electrically conducting material, such as Gold (AuAu) or Platinum (PtPt).

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A vacuum environment is essential because air molecules would scatter the electron beam, preventing it from focusing and reaching the sample.

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The resolution of an SEM can range from 1 nm1\text{ nm} to 20 nm20\text{ nm}, which is significantly better than the 200 nm200\text{ nm} limit of conventional light microscopes.

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The magnification (MM) in an SEM is the ratio of the length of the scan on the display monitor to the length of the scan on the specimen.

📐Formulae

M=LdisplayLspecimenM = \frac{L_{display}}{L_{specimen}}

λ≈1.23V nm (where V is the accelerating voltage in volts)\lambda \approx \frac{1.23}{\sqrt{V}} \text{ nm (where } V \text{ is the accelerating voltage in volts)}

R=0.61λNA (Rayleigh Criterion for resolution limit)R = \frac{0.61 \lambda}{NA} \text{ (Rayleigh Criterion for resolution limit)}

💡Examples

Problem 1:

An electron microscope uses an accelerating voltage of 10,000 V10,000\text{ V}. Calculate the approximate de Broglie wavelength (λ\lambda) of the electrons used. If a light microscope uses blue light with λ=400 nm\lambda = 400\text{ nm}, how many times smaller is the electron wavelength?

Solution:

Using the simplified formula for electron wavelength: λe≈1.2310000 nm\lambda_{e} \approx \frac{1.23}{\sqrt{10000}} \text{ nm} λe≈1.23100 nm=0.0123 nm\lambda_{e} \approx \frac{1.23}{100} \text{ nm} = 0.0123 \text{ nm} To find the ratio: Ratio=λlightλe=4000.0123≈32520\text{Ratio} = \frac{\lambda_{light}}{\lambda_{e}} = \frac{400}{0.0123} \approx 32520

Explanation:

The wavelength of the electron at 10,000 V10,000\text{ V} is approximately 0.0123 nm0.0123\text{ nm}. This wavelength is over 32,00032,000 times smaller than visible blue light, which explains why SEM can achieve much higher resolution.

Problem 2:

A scientist is observing a pollen grain. The scan length on the specimen is 10 \mum10 \text{ \mu m} (10×10−6 m10 \times 10^{-6} \text{ m}), and the resulting image on the screen is 5 cm5 \text{ cm} (0.05 m0.05 \text{ m}) wide. Calculate the magnification (MM).

Solution:

The formula for magnification is: M=Image SizeObject SizeM = \frac{\text{Image Size}}{\text{Object Size}} Convert both to the same units (meters): Image size=0.05 m\text{Image size} = 0.05 \text{ m} Object size=10×10−6 m\text{Object size} = 10 \times 10^{-6} \text{ m} M=0.0510×10−6M = \frac{0.05}{10 \times 10^{-6}} M=5×10−210×10−6M = \frac{5 \times 10^{-2}}{10 \times 10^{-6}} M=0.5×104=5,000M = 0.5 \times 10^{4} = 5,000

Explanation:

By dividing the physical size of the image on the display by the actual area scanned on the specimen, we find the magnification is 5,000×5,000\times.

Scanning Electron Microscope (SEM)-advanced Class 9 Notes & Examples