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 least?
(a) For the electron,
E = 3.2*10^-18 J
momentum = sqrt(2m*E)
= 2.41*10^-24 kg m/s
wavelength = h/(momentum) = 2.74*10^-10 m
For the photon, the energy is the same but
wavelength = h*c/(Energy)
= 6.2*10^-8 m
(b) Use the formulas
E(photon) = h*c/(wavelength)
E(electron) = (momentum)^2/(2*Mp)
= [h/(wavelength)]^2/(2*Mp)
where Mp is the proton mass.
(c) Pick the microscope that uses particles of least energy, for a given wavelength. The wavelength determines the resolving power.
Wavelength and resolution element size are roughly the same. Look up the appropriate formulas online.
a) Well, a photon and an electron walk into a bar...oh wait, you're looking for wavelengths, not jokes! My bad. The wavelength of a photon can be calculated using the formula λ = hc/E, where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and E is the energy. Plugging in the values, we get λ = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (20.0 eV * 1.6 x 10^-19 J/eV). Crunching the numbers, the wavelength of the photon is approximately 619 nm.
b) Alright, let's move on to the energy wavelength duo! To calculate the energy of a photon, you can use the formula E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the values, we have E = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (250 nm * 10^-9 m/nm). After some calculations, the energy of the photon is around 7.96 eV.
c) Ah, the serious part! If you want to study an organic molecule that is about 250 nm long, it's best to use a wavelength close to that size. So, the correct wavelength would be around 250 nm, as you mentioned. As for the probe that will likely damage the molecule the least, I'd have to say go with the photon. Remember, photons are usually less massive and have less interaction with the sample compared to electrons. So, go with the photon and give that molecule some space to breathe without causing much damage!
a) To find the wavelength of a photon with energy E, we can use the formula:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength.
Rearranging the formula to solve for wavelength, we have:
λ = hc/E
Substituting the values, we get:
λ = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (20.0 eV * 1.602 x 10^-19 J/eV)
Simplifying this expression, we find:
λ ≈ 9.86 x 10^-8 m
Therefore, the wavelength of the photon with an energy of 20.0 eV is approximately 9.86 x 10^-8 m.
To find the wavelength of an electron with the same energy, we can use de Broglie's equation:
λ = h / p
where λ is the wavelength, h is Planck's constant, and p is the momentum of the electron.
The momentum of an electron can be calculated using the relativistic energy-momentum equation:
E = (m^2 * c^4 + p^2 * c^2)^(1/2)
where E is the energy, m is the rest mass of the electron (9.11 x 10^-31 kg), and c is the speed of light.
In this case, the energy of the electron is given as 20.0 eV. Converting it to joules:
E = 20.0 eV * 1.602 x 10^-19 J/eV
Substituting this into the relativistic energy-momentum equation, we get:
20.0 eV * 1.602 x 10^-19 J/eV = (9.11 x 10^-31 kg)^2 * c^4 + p^2 * (3.00 x 10^8 m/s)^2
Simplifying the equation, we can solve for the momentum:
p ≈ 2.47 x 10^-24 kg m/s
Now, substituting this momentum into de Broglie's equation, we can find the wavelength:
λ = (6.626 x 10^-34 J*s) / (2.47 x 10^-24 kg m/s)
Simplifying this expression, we find:
λ ≈ 2.68 x 10^-10 m
Therefore, the wavelength of the electron with an energy of 20.0 eV is approximately 2.68 x 10^-10 m.
b) To find the energy of a photon with wavelength λ, we can use the formula:
E = hc/λ
where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength.
Substituting the given wavelength of 250 nm (or 250 x 10^-9 m), we get:
E = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (250 x 10^-9 m)
Simplifying this expression, we find:
E ≈ 7.95 x 10^-19 J
Therefore, the energy of the photon with a wavelength of 250 nm is approximately 7.95 x 10^-19 J.
To find the energy of an electron with the same wavelength, we can use de Broglie's equation:
λ = h / p
where λ is the wavelength, h is Planck's constant, and p is the momentum of the electron.
The momentum of an electron can be calculated using the relation:
p = mv
where m is the mass of the electron (9.11 x 10^-31 kg) and v is the speed of the electron.
To calculate v, we can use the formula:
v = c * (λ / c)
where c is the speed of light.
Substituting the given wavelength of 250 nm (or 250 x 10^-9 m), and the speed of light, we get:
v = 3.00 x 10^8 m/s * (250 x 10^-9 m / 3.00 x 10^8 m/s)
Simplifying this expression, we find:
v ≈ 2.50 x 10^2 m/s
Now, substituting the mass and speed into the momentum formula, we get:
p = (9.11 x 10^-31 kg) * (2.50 x 10^2 m/s)
Simplifying this expression, we find:
p ≈ 2.28 x 10^-28 kg m/s
Now, substituting the momentum into de Broglie's equation, we can find the energy:
E = (6.626 x 10^-34 J*s) / (2.28 x 10^-28 kg m/s)
Simplifying this expression, we find:
E ≈ 2.90 x 10^-6 J
Therefore, the energy of the electron with a wavelength of 250 nm is approximately 2.90 x 10^-6 J.
c) When studying an organic molecule that is about 250 nm long, it is better to use a photon microscope with a wavelength close to the size of the molecule. In this case, using a wavelength of around 250 nm would be appropriate.
Using a photon microscope with this wavelength is likely to cause less damage to the organic molecule. Electrons have a higher energy and can damage or alter the molecules they interact with due to their higher interaction strength. Photons, on the other hand, have less energy and are less likely to cause damage to the organic molecule during the imaging process.
To answer these questions, we need to use the formulas related to the energy and wavelength of photons and electrons.
a) To find the wavelength of a photon or an electron with a given energy, we can use the equation:
Energy (E) = Planck's constant (h) * speed of light (c) / wavelength (λ)
Given that the energy is the same for both the photon and the electron (20.0 eV), we can equate the equations for the photon and the electron:
For the photon:
20.0 eV = h * c / λ_photon
For the electron:
20.0 eV = (h^2 / (2 * m * e)) / λ_electron
Where m is the mass of the electron and e is the elementary charge.
Since we are looking for the wavelengths, we can rearrange both equations to solve for λ_photon and λ_electron respectively:
For the photon:
λ_photon = h * c / 20.0 eV
For the electron:
λ_electron = h^2 / (2 * m * e * 20.0 eV)
b) To find the energy of a photon or an electron with a given wavelength, we can rearrange the above equation to solve for energy (E):
For the photon:
E_photon = h * c / λ_photon
For the electron:
E_electron = (h^2 / (2 * m * e)) / λ_electron
c) To determine the ideal wavelength and less damaging probe, we need to consider the wavelength range and interaction mechanisms of photons and electrons with the organic molecule.
Photon microscopy, such as optical microscopy, typically operates in the visible to near-infrared range (approximately 400 nm to 700 nm). As the molecule is about 250 nm long, using a photon microscope in this range would be suitable. Photons interact with molecules primarily through absorption or scattering, which generally does not cause significant damage.
Electron microscopy, such as transmission electron microscopy (TEM), operates at much smaller wavelengths (approximately 0.1 nm to 10 nm). The wavelength of 250 nm is considerably larger than the range of electron microscopy. Electrons interact with molecules through a process called electron scattering, which can cause damage due to their higher energy and smaller wavelength. The electron microscope is more likely to damage the organic molecule compared to the photon microscope.
Therefore, to study the organic molecule that is approximately 250 nm long, it is advisable to use a photon microscope with a wavelength range around 400 nm to 700 nm, as it is less likely to damage the molecule.