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(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 an electron each have the same wavelength of 250 nm, find the energy of each. (c) You want to study an organic molecule that is about 250 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?

Short Answer

Expert verified
(a) Photon: 62.06 nm, Electron: 2.73 nm. (b) Photon: 4.97 eV, Electron: 0.015 eV. (c) Use 250 nm wavelength; electrons are less damaging.

Step by step solution

01

Wavelength of Photon for 20.0 eV

To find the wavelength of a photon with energy 20.0 eV, use the formula: \[ \lambda = \frac{hc}{E} \]where:- \( h = 6.626 \times 10^{-34} \) J\cdot s (Planck's constant)- \( c = 3.00 \times 10^8 \) m/s (speed of light)- \( E = 20.0 \times 1.602 \times 10^{-19} \) J (energy converted from eV to Joules)Calculating this gives:\[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{20.0 \times 1.602 \times 10^{-19}} \approx 62.06 \ nm \]
02

Wavelength of Electron for 20.0 eV

To find the wavelength of an electron with energy 20.0 eV, use the de Broglie wavelength formula: \[ \lambda = \frac{h}{\sqrt{2mE}} \]where:- \( m = 9.11 \times 10^{-31} \) kg (mass of electron)- \( E = 20.0 \times 1.602 \times 10^{-19} \) JCalculating this gives:\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 20.0 \times 1.602 \times 10^{-19}}} \approx 2.73 \ nm \]
03

Energy of Photon for 250 nm

For a photon with a wavelength of 250 nm, use the formula:\[ E = \frac{hc}{\lambda} \]where \( \lambda = 250 \times 10^{-9} \) m.Calculating this gives:\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{250 \times 10^{-9}} \approx 4.97 \ eV \]
04

Energy of Electron for 250 nm

Using the de Broglie relation \( \lambda = \frac{h}{\sqrt{2mE}} \), solve for \( E \) when \( \lambda = 250 \times 10^{-9} \) m.Square both sides to isolate \( E \):\[ E = \frac{h^2}{2m\lambda^2} \]Calculating this gives:\[ E = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (250 \times 10^{-9})^2} \approx 1.5 \times 10^{-2} \ eV \]
05

Wavelength for Studying a Molecule and Choosing a Probe

For studying an organic molecule approximately 250 nm long, use a wavelength around 250 nm for better resolution. From previous computations, an electron with 250 nm wavelength has significantly lower energy (\(1.5 \times 10^{-2} \) eV) compared to a photon (4.97 eV). Thus, an electron beam is less likely to damage the molecule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

De Broglie Wavelength
The concept of De Broglie wavelength helps us understand the wave nature of particles like electrons. The De Broglie wavelength is given by the formula \[ \lambda = \frac{h}{\sqrt{2mE}} \]where:- \( \lambda \) is the wavelength,- \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\),- \( m \) is the mass of the particle (for an electron, \(9.11 \times 10^{-31} \text{ kg}\)),- \( E \) is the kinetic energy of the particle.

In the original exercise, this formula is used to calculate the wavelength of an electron with a given energy. The beauty of the De Broglie wavelength is its ability to describe wave-like properties in small particles, linking energy with wavelength in a tangible manner. This relationship is pivotal in fields such as electron microscopy, where understanding wavelength is crucial.
  • Wave nature of particles: Describes how particles like electrons exhibit wave characteristics.
  • Central to quantum mechanics: Foundation for understanding phenomena at atomic scales.
Photon Energy
Photons, the basic units of light, show unique properties as they behave like both waves and particles. The energy \( E \) of a photon is associated with its wavelength by the formula:\[ E = \frac{hc}{\lambda} \]Here:- \( E \) is the energy of the photon,- \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J} \cdot \text{s})\),- \( c \) is the speed of light \((3.00 \times 10^{8} \text{ m/s})\),- \( \lambda \) is the wavelength in meters.

For example, a photon with a wavelength of 250 nm has an energy of approximately 4.97 eV. This relationship is fundamental when examining photons' interactions with matter, as their energy is a critical factor.
  • Determines how light interacts with materials.
  • Helps to calculate wavelength from energy or vice versa.
  • Fundamental to technologies like solar cells and spectroscopy.
Electron Microscopy
Electron microscopy is a powerful technique used to visualize tiny objects with high precision. It leverages the wave nature of electrons to achieve superior resolution compared to light microscopy. It functions by using accelerated electrons with wavelengths calculated by the De Broglie relation.

This technology allows scientists to study structures as small as individual atoms, making it indispensable in fields like material science, nanotechnology, and biology.

Choosing the right probe is crucial, especially for delicate samples, like organic molecules. In the exercise, the electron beam with lower energy than photons causes less damage, making it the preferred choice for studying fragile organic structures.
  • Reveals structural details: Provides insight into molecular and atomic architecture.
  • Ultrafine resolution: Unmatched in revealing fine structures at microscopic levels.
  • Low-energy probes for delicate samples: Preferred method to prevent sample damage.
Organic Molecule Study
Studying organic molecules often requires a careful balance between resolution and sample integrity. In the scenario of observing a molecule approximately 250 nm long, it is essential to select a technique that provides precise imaging without altering the sample's structure.

Opting for a wavelength close to the size of the object yields better resolution, which is why in this exercise, a 250 nm wavelength is used. However, photons at this wavelength have higher energy that can damage the sample.

An electron microscope, using low-energy electrons, ensures minimal damage while still achieving high-resolution images.
  • A delicate balance: Choose wavelengths that optimize resolution without causing damage.
  • Essential in research: Critical for studying biological and chemical properties.

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Most popular questions from this chapter

(a) What is the energy of a photon that has wavelength 0.10 \(\mu\)m ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 \(\mu\)m in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

(a) For one-electron ions with nuclear charge Z, what is the speed of the electron in a Bohr-model orbit labeled with \(n\)? Give your answer in terms of \(v_1\), the orbital speed for the \(n\) = 1 Bohr orbit in hydrogen. (b) What is the largest value of Z for which the \(n\) = 1 orbital speed is less than 10\(\%\) of the speed of light in vacuum?

An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy \(K\) of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron?

How does the wavelength of a helium ion compare to that of an electron accelerated through the same potential difference? (a) The helium ion has a longer wavelength, because it has greater mass. (b) The helium ion has a shorter wavelength, because it has greater mass. (c) The wavelengths are the same, because the kinetic energy is the same. (d) The wavelengths are the same, because the electric charge is the same.

The wavelength 10.0 \(\mu\)m is in the infrared region of the electromagnetic spectrum, whereas 600 nm is in the visible region and 100 nm is in the ultraviolet. What is the temperature of an ideal blackbody for which the peak wavelength \(\lambda_m\) is equal to each of these wavelengths?

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