(a) If a photon and an electron each have the same energy of...

(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and anelectron each have the same wavelength of 250 $\mathrm{nm}$ , find the energy of each. (c) You want to study an organic molecule that is about 250 $\mathrm{nm}$ long using either a photon or an electron microscope .Approximately what wavelength should you use, and which probe, the electron or the photon, is likely to damage the molecule the least?

(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and anelectron each have the same wavelength of 250 $\mathrm{nm}$ , find the energy
of each. (c) You want to study an organic molecule that is about 250 $\mathrm{nm}$ long using either a photon or an electron microscope .Approximately what wavelength should you use, and which probe,
the electron or the photon, is likely to damage the molecule the least?

Key Concepts

Probe-Induced Sample Damage

Probe-induced sample damage refers to the detrimental effects that can occur when a high-energy imaging probe interacts with a specimen. In microscopy, different probes such as photons and electrons have varying levels of energy and interactions with matter, so understanding the potential for damage is crucial when imaging sensitive samples. The choice between a lower-energy probe that minimizes damage and a higher-energy one that provides better resolution is a key consideration in experimental design.

Photon Energy-Wavelength Relationship

This concept describes the direct connection between a photon's energy and its wavelength, defined by the equation E = hc/?, where E is energy, h is Planck’s constant, and c is the speed of light. It is fundamental in quantum physics and optics, linking the quantized nature of electromagnetic radiation to its observed wavelength and frequency.

de Broglie Wavelength

The de Broglie wavelength concept states that all matter exhibits wave-like properties, with a wavelength given by ? = h/p, where h is Planck’s constant and p is the momentum of the particle. For particles like electrons, especially when their kinetic energy is known, this relationship allows for the calculation of their wavelength, which is essential for understanding quantum behavior and interference phenomena.

Diffraction Limit and Microscopy Resolution

The diffraction limit is a principle in optics and microscopy that relates the resolving power of an imaging system to the wavelength of the probe used. A smaller wavelength allows for finer resolution of structures. This concept is vital when choosing between different types of microscopes, as it connects the achievable detail in the image to the wavelength of the photons or electrons used in the probe.

Transcript

00:01 Now we are given for the first part of this question, a photon and an electron having the same energy.

00:12 Turning ev, you know to find what is the wavelength.

00:17 So for the photon, energy can be written as hc over lambda.

00:28 So its wavelength is just hc over the energy, where hc can use a...

00:40 1240 ev dot nanometers and the energy is 20 ev so we should get 62 nanometers for the wavelength of the photon now for the electron on the other head this energy is related to its kinetic energy right this equals to its momentum right momentum squared over 2m b squared over 2m and we gotta use this to find what is our nebrokly's wavelength given as h over p which would be equals to h over square root of 2m times the kinetic energy now we will have to convert the 20 ev into jules for this by multiplying by 1 .6 times 10 power minus 90.

02:27 Then we should get a wavelength of about 2 .7 .4 times 10 power minus 10 meters or just 0 .274 nanometers.

02:47 For the next part of this question, we are given both of them have the same wavelength, both the photon and the electron have wavelength of 250 nanometers and to find what are their various energies so for the photon is very simple energy is equals to hc over lambda again we use h times c to be 1240 ev.

03:37 Then divided by 250 we should get 4 .96 electron for the electron the wavelength signifies to us what is the momentum which is about 2 .65 times 10 power minus 27 from this we can calculate the energy which is just p square 2m this should get us about 3 .8, 5 .6 times 10 power minus 24 joules.

04:59 But in order to compare it with our photon, we have to convert this into electron volts.

05:08 So we divide this by 1 .6 10 power minus 19...

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