Question

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

Short answer

Light photon energy: 248 eV; electron energy: 0.375 eV. Electrons are preferred, causing less damage.

Step-by-step solution

Calculate photon energy using light

To find the energy of a photon, we use the equation $E=\frac{hc}{\lambda }$, where $h$ is Planck's constant $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$, $c$ is the speed of light $3.00\times {10}^8\text{ m/s}$, and $\lambda =5.00\times {10}^{-9}\text{ m}$ is the wavelength. Calculate $E$:$$E=\frac{6.626\times {10}^{-34}\times 3.00\times {10}^8}{5.00\times {10}^{-9}}=3.97\times {10}^{-17}\text{ J}$$Convert the energy from Joules to electronvolts (1 eV = $1.602\times {10}^{-19}\text{ J}$):$$E=\frac{3.97\times {10}^{-17}}{1.602\times {10}^{-19}}=248\text{ eV}$$

Calculate kinetic energy of an electron using matter waves

To find the kinetic energy of an electron with a given wavelength, use the de Broglie wavelength formula $\lambda =\frac{h}{p}$ where $p$ is momentum. Hence the momentum $p=\frac{h}{\lambda }$. Since $p=mv$ and kinetic energy (KE) is $\frac{1}{2}m{v}^2$, we can express $KE$ as $\frac{p^2}{2m}$:$$KE=\frac{h^2}{2m{\lambda }^2}$$Here, $m=9.11\times {10}^{-31}\text{ kg}$ (electron mass) and $\lambda =5.00\times {10}^{-9}\text{ m}$.$$KE=\frac{(6.626\times {10}^{-34}{)}^2}{2\times 9.11\times {10}^{-31}\times (5.00\times {10}^{-9}{)}^2}=6.02\times {10}^{-20}\text{ J}$$Convert to eV:$$KE=\frac{6.02\times {10}^{-20}}{1.602\times {10}^{-19}}=0.375\text{ eV}$$

Compare photon and electron energy

The energy of a photon with a 5 nm wavelength is 248 eV, while the kinetic energy of an electron with the same wavelength is 0.375 eV. This significant difference in energy levels explains why matter waves (like those in electron microscopes) are preferable, as they provide similar wavelengths for imaging while causing less potential damage.

Key concepts

de Broglie wavelength

The de Broglie wavelength is a fundamental concept in physics that helps us understand the wave-particle duality of matter.
Louis de Broglie proposed that particles such as electrons exhibit both particle and wave properties, which can be described by a wavelength.
This wavelength is known as the de Broglie wavelength, and it is extremely useful in electron microscopy.
To calculate the de Broglie wavelength ($\lambda$) of a particle, you use the equation $\lambda =\frac{h}{p}$, where $h$ is Planck's constant, and $p$ is the momentum of the particle. Momentum $p$ is the product of mass ($m$) and velocity ($v$) of the particle.

• Momentum is given by $p=mv$
• Planck's constant is $6.626\times {10}^{-34}\text{J}\cdot \text{s}$

The de Broglie wavelength is particularly important when it comes to using electron microscopes.
Electrons with a smaller wavelength have higher resolution, allowing us to see extremely small details that are important for studying tiny objects like viruses.

photon energy

Photon energy is crucial when dealing with electromagnetic waves, such as light.
Photons, the basic unit of light, carry energy that can be determined using the wavelength of the light.
The energy $E$ of a photon can be calculated using the formula $E=\frac{hc}{\lambda }$.

• Here, $h$ is the 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 of the photon

When comparing photons with different wavelengths, shorter wavelengths result in higher photon energy.
This is especially important in imaging, where high-energy photons can damage delicate structures, like viruses, which is why alternative methods, such as electron matter waves, are sometimes preferred.

kinetic energy of electrons

The kinetic energy of electrons is another essential aspect when we study tiny objects using electron beams.
Unlike photons, electrons have mass, and their kinetic energy is directly related to their momentum and speed.
The kinetic energy ($KE$) of an electron can be determined from its momentum ($p$) using the relation $KE=\frac{p^2}{2m}$, where $m$ is the mass of the electron.

• Momentum can be expressed using the de Broglie relation: $p=\frac{h}{\lambda }$
• The mass of an electron is $9.11\times {10}^{-31}\text{kg}$

This formula shows that by knowing the wavelength $\lambda$ of an electron, one can determine its kinetic energy.
A key point in electron microscopy is that electrons at the same wavelength as photons have dramatically lower kinetic energies.
This lowers the risk of damaging the sample while still providing a high-resolution image, crucial for delicate samples like microorganisms.