Answer :
To address this problem, we need to determine whether relativistic equations are necessary and calculate the width of the slit responsible for the electron diffraction pattern.
(a) Do we need to use relativity formulas?
To determine if relativistic effects are significant, we need to calculate the speed of the electrons after being accelerated through a potential difference of 0.100 kV (100 V). Using non-relativistic physics, the kinetic energy gained by the electron is equal to the electric potential energy:
[tex]KE = eV[/tex]
Where:
- [tex]e[/tex] is the charge of the electron (approximately [tex]1.602 \times 10^{-19}[/tex] C).
- [tex]V[/tex] is the potential difference, which is 0.100 kV or 100 V.
This kinetic energy results in a velocity [tex]v[/tex] such that:
[tex]\frac{1}{2}mv^2 = eV[/tex]
Rearranging for [tex]v[/tex]:
[tex]v = \sqrt{\frac{2eV}{m}}[/tex]
Where [tex]m[/tex] is the mass of an electron (approximately [tex]9.109 \times 10^{-31}[/tex] kg).
Substitute known values:
[tex]v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \times 100}{9.109 \times 10^{-31}}} \approx 5.93 \times 10^6 \text{ m/s}[/tex]
Compare this speed to the speed of light [tex]c \approx 3 \times 10^8[/tex] m/s. Since [tex]v \ll c[/tex], relativistic effects are negligible, and non-relativistic formulas are sufficient.
(b) How wide is the slit?
The first diffraction minima for electrons passing through a slit occurs at an angle [tex]\theta[/tex] given by the formula:
[tex]a \sin \theta = m \lambda[/tex]
Where:
- [tex]a[/tex] is the slit width we need to calculate.
- [tex]\theta = 11.5^\circ[/tex] for the first minima.
- [tex]m = 1[/tex] since we're considering the first minima.
- [tex]\lambda[/tex] is the wavelength of the electron, given by the de Broglie wavelength formula:
[tex]\lambda = \frac{h}{mv}[/tex]
Using Planck's constant [tex]h = 6.626 \times 10^{-34}[/tex] J·s, calculate [tex]\lambda[/tex]:
[tex]\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 5.93 \times 10^6} \approx 1.23 \times 10^{-10} \text{ m}[/tex]
Substitute [tex]\lambda[/tex], [tex]\theta[/tex], and [tex]m[/tex] into the diffraction formula:
[tex]a \sin(11.5^\circ) = 1 \times 1.23 \times 10^{-10}[/tex]
Calculate [tex]a[/tex]:
[tex]a = \frac{1.23 \times 10^{-10}}{\sin(11.5^\circ)} \approx 6.15 \times 10^{-10} \text{ m}[/tex]
Thus, the width of the slit is approximately [tex]6.15 \times 10^{-10}[/tex] meters.
