To what voltages must we accelerate electrons (as in an electron microscope, for example) if we
wish to resolve;
a) a virus of diameter 12 nm , b) an atom of 0.12 nm, c) a proton of diameter 1.2 fm?
The formula you need can be found here:
http://www.ou.edu/research/electron/bmz5364/resolutn.html
To calculate the voltage required to resolve each of the given particles, we can use the de Broglie wavelength equation, which relates the wavelength of an object to its momentum:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J.s), and p is the momentum.
The momentum of a particle can be calculated using the equation:
p = mv
where m is the mass of the particle and v is its velocity.
Let's calculate the voltage required for each case:
a) For a virus with a diameter of 12 nm:
The mass of a virus can vary widely, but let's assume a mass of 1 x 10^-15 kg (typical mass for a virus). To resolve the virus, we need to ensure that the de Broglie wavelength is smaller than the virus diameter.
λ = h / p
λ = h / (mv)
λ = h / (m * v)
Since the virus is stationary (v = 0), we can calculate the upper limit of the wavelength:
λ = h / (m * v)
λ = h / (m * 0)
λ = ∞
As the de Broglie wavelength is infinite when the virus is stationary, we can't resolve it using electron microscopy.
b) For an atom with a diameter of 0.12 nm:
The mass of an atom can also vary based on the element, but let's assume a mass of 1.67 x 10^-27 kg (mass of a hydrogen atom).
λ = h / p
λ = h / (mv)
Using the de Broglie wavelength equation, we can calculate the required velocity to resolve the atom:
λ = h / (m * v)
v = h / (m * λ)
Substituting the values:
v = (6.626 x 10^-34 J.s) / (1.67 x 10^-27 kg * 0.12 x 10^-9 m)
v ≈ 2.5 x 10^6 m/s
To calculate the required voltage, we can use the equation for the kinetic energy of an electron:
E = (1/2)mv²
Simplifying and substituting the values:
E = (1/2) * (9.11 x 10^-31 kg) * (2.5 x 10^6 m/s)²
E ≈ 2.26 x 10^-17 J
Finally, we can use the equation for the electric potential energy of an electron:
E = qV
Solving for V:
V = E / q
V = (2.26 x 10^-17 J) / (1.6 x 10^-19 C)
V ≈ 141.25 V
Therefore, in order to resolve an atom with a diameter of 0.12 nm, we would need to accelerate the electrons to a voltage of approximately 141.25 V.
c) For a proton with a diameter of 1.2 fm:
For the proton, we can perform a similar calculation.
λ = h / p
λ = h / (mv)
Substituting the values:
λ = (6.626 x 10^-34 J.s) / [(1.67 x 10^-27 kg) * (v)]
To resolve the proton, the de Broglie wavelength should be smaller than the proton diameter:
λ ≤ diameter
λ ≤ 1.2 fm
Thus, we can calculate the upper limit for the velocity:
(6.626 x 10^-34 J.s) / [(1.67 x 10^-27 kg) * (v)] ≤ 1.2 fm
Simplifying:
v ≥ [(6.626 x 10^-34 J.s) / (1.67 x 10^-27 kg)] / (1.2 x 10^-15 m)
v ≥ 0.51 x 10^6 m/s
The required voltage can be determined in the same way as for the atom:
E = (1/2)mv²
Simplifying and substituting the values:
E = (1/2) * (9.11 x 10^-31 kg) * (0.51 x 10^6 m/s)²
E ≈ 1.17 x 10^-19 J
V = E / q
V = (1.17 x 10^-19 J) / (1.6 x 10^-19 C)
V ≈ 0.73 V
Therefore, to resolve a proton with a diameter of 1.2 fm, we need to accelerate the electrons to a voltage of approximately 0.73 V.
To determine the voltages required to resolve different objects using an electron microscope, we can use the concept of wavelength resolution. The resolving power of an electron microscope is governed by the de Broglie wavelength of electrons, which depends on their acceleration voltage. The de Broglie wavelength of an electron is given by the equation:
λ = h / √(2mE)
Where:
λ is the de Broglie wavelength of the electron,
h is Planck's constant (6.626 x 10^-34 Js),
m is the mass of the electron (9.109 x 10^-31 kg),
E is the kinetic energy of the electron.
To resolve objects of different sizes, we need to find the appropriate de Broglie wavelength and then calculate the required acceleration voltage.
a) Virus with diameter 12 nm:
The de Broglie wavelength necessary to resolve the virus can be calculated using the equation above. We assume the virus as a spherical object, thus the de Broglie wavelength should be at least twice the size of the virus for proper resolution.
λ = 2 * diameter = 2 * (12 x 10^-9 m) = 24 x 10^-9 m
Now we can rearrange the equation to solve for the acceleration voltage E:
E = h^2 / (2mλ^2)
Plugging in the values:
E = (6.626 x 10^-34 Js)^2 / (2 * (9.109 x 10^-31 kg) * (24 x 10^-9 m)^2)
Calculating this equation gives us the required acceleration voltage.
b) Atom with diameter 0.12 nm:
Following the same procedure as above, we calculate the de Broglie wavelength for an atom of 0.12 nm in diameter:
λ = 2 * diameter = 2 * (0.12 x 10^-9 m) = 0.24 x 10^-9 m
Now we can plug in this value into the acceleration voltage equation and solve for E.
c) Proton with diameter 1.2 fm:
The diameter of a proton is extremely small, so we'll use femtometers (fm) for calculations instead of nanometers (nm). Following the same steps as before, we calculate the de Broglie wavelength for a proton of diameter 1.2 fm:
λ = 2 * diameter = 2 * (1.2 x 10^-15 m) = 2.4 x 10^-15 m
Plug this value into the acceleration voltage equation to find the required E.
By calculating these equations, you can determine the acceleration voltages needed to resolve the specified objects in an electron microscope.