To determine the kinetic energy an electron must have to resolve a protein molecule with a diameter of 8.60 nm, we need to relate the wavelength of the electron to the size of the object we want to observe. This relationship is grounded in the principles of wave-particle duality, which states that particles like electrons exhibit both particle-like and wave-like properties.
Understanding the Relationship Between Wavelength and Kinetic Energy
The wavelength (\( \lambda \)) of an electron can be calculated using the de Broglie wavelength formula:
\( \lambda = \frac{h}{p} \)
Here, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), and \( p \) is the momentum of the electron. The momentum can be expressed as:
\( p = mv \)
where \( m \) is the mass of the electron and \( v \) is its velocity. The kinetic energy (\( KE \)) of the electron is given by:
\( KE = \frac{1}{2} mv^2 \)
Connecting Kinetic Energy to Wavelength
To find the kinetic energy in terms of wavelength, we can rearrange the de Broglie equation. First, substituting \( p \) into the wavelength equation gives:
\( \lambda = \frac{h}{mv} \)
From this, we can express the velocity \( v \) as:
\( v = \frac{h}{m\lambda} \)
Now, substituting this expression for \( v \) back into the kinetic energy formula yields:
\( KE = \frac{1}{2} m \left(\frac{h}{m\lambda}\right)^2 \)
This simplifies to:
\( KE = \frac{h^2}{2m\lambda^2} \)
Calculating the Required Kinetic Energy
Now, we can plug in the values. First, we need to convert the diameter of the protein molecule from nanometers to meters:
\( 8.60 \, \text{nm} = 8.60 \times 10^{-9} \, \text{m} \)
Next, substituting the known values into the kinetic energy formula:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Wavelength \( \lambda = 8.60 \times 10^{-9} \, \text{m} \)
Now, substituting these values into the equation:
\( KE = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (8.60 \times 10^{-9})^2} \)
Calculating the numerator:
\( (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2 \)
Calculating the denominator:
\( 2 \times (9.11 \times 10^{-31}) \times (8.60 \times 10^{-9})^2 = 2 \times (9.11 \times 10^{-31}) \times (7.396 \times 10^{-17}) = 1.34 \times 10^{-47} \, \text{kg m}^2 \)
Now, substituting these values back into the kinetic energy equation:
\( KE = \frac{4.39 \times 10^{-67}}{1.34 \times 10^{-47}} \approx 3.27 \times 10^{-20} \, \text{J} \)
Converting to Electronvolts
Since kinetic energy is often expressed in electronvolts (eV), we can convert joules to eV using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):
\( KE \approx \frac{3.27 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.204 \, \text{eV} \)
Thus, the kinetic energy required for an electron to resolve a protein molecule that is 8.60 nm in diameter is approximately 0.204 eV. This energy allows the electron's wavelength to be comparable to the size of the protein, enabling effective imaging in an electron microscope.
