To resolve details of an object, you must use a wavelength...

To resolve details of an object, you must use a wavelength that is about the same size, or smaller, than the details you want to observe. Suppose you want to study a molecule that is about $1.000 \times 10^{-10} \mathrm{~m}$ in length. (a) What minimum photon energy is required to study this molecule? (b) What is the minimum kinetic energy of electrons that could study this? (c) Through what potential difference should the electrons be accelerated to reach this enerov?

To resolve details of an object, you must use a wavelength that is about the same size, or smaller, than the details you want to observe. Suppose you want to study a molecule that is about $1.000 \times 10^{-10} \mathrm{~m}$ in length.
(a) What minimum photon energy is required to study this molecule? (b) What is the minimum kinetic energy of electrons that could study this? (c) Through what potential difference should the electrons be accelerated to reach this enerov?

Key Concepts

Diffraction Limit and Resolution

This concept pertains to the fact that to observe or resolve an object, the wavelength of the probing radiation must be comparable to or smaller than the size of the object's features. It sets a fundamental limit on the resolution in imaging systems and is crucial in fields ranging from microscopy to astronomy.

Photon Energy and Wavelength

A central concept in quantum mechanics is that the energy of a photon is inversely proportional to its wavelength, expressed by the equation E = hc/?, where h is Planck's constant and c is the speed of light. This relation shows that shorter wavelengths are associated with higher energy photons, which is essential for resolving very small details in a specimen.

de Broglie Wavelength

Introduced by Louis de Broglie, this concept extends the wave-particle duality to matter particles such as electrons. It states that every particle has an associated wavelength given by ? = h/p, where p is the momentum. This principle is used to determine the necessary kinetic energy of electrons to achieve a de Broglie wavelength small enough to resolve molecular or atomic structures.

Electron Acceleration and Potential Difference

In an electron accelerator or electron microscope, electrons gain kinetic energy when they are accelerated through a potential difference. The acquired kinetic energy is equivalent to the electrical potential energy (eV), where e is the charge of the electron and V is the potential difference. This relationship directly links the accelerating voltage to the electrons’ capability to achieve a desired de Broglie wavelength for imaging.

Transcript

00:02 Hello, and in this question here, we want to resolve an object, which is 10 to the power of minus 10 meters in length, okay? and we want to know what the kinetic energy that we would need to, the kinetic energy we would need if we were to do this with a photon and with an electron.

00:22 Okay? so, if the object we want to resolve is around 10 to the power of minus 10, this means the wavelength of the photon, or the electron, so the debroider wavelength, for the case of the electron, must be in the order of 10 to the power of minus 10.

00:40 So we have the, so if we start off first with the photon, the energy of a photon, okay, is equal to planck's constant times the speed of light, sorry, times the frequency.

00:55 Okay? and the frequency is equal to the speed of light divided by the wavelength.

01:02 So we get to the energy of the photon is equal to the speed of life divided by the wavelength.

01:08 Now, i've already said in order to resolve an object that is 10 to the power of minus 10, we'd want the wavelength of the photon to be equal to 10 to the power of minus 10 meters.

01:19 Okay? so this means we know what the wavelength is.

01:25 The speed of light, well, that's a constant, and that is equal to 3 times 10 to the power of 8 meters per second.

01:31 And planx constant is equal to 6 .624 times 10 to the power of minus 34 with units of, well, it would be joules per second.

01:50 Am i correct? no, dual seconds.

01:54 Jewel seconds.

01:57 Okay.

01:59 So, we can fill these values in and get that the energy of the photon is equal to 1 .9.

02:07 98, 7 -8 times 10 to the power of minus 15 joules.

02:14 Okay? and that there is our energy of our photon.

02:18 Now, if we're going to repeat this for an electron, okay? well, we want to find the kinetic energy of the electron when the de broilie wavelength is 10 to the power of minus 10...

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