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.