Answer:
Explanation:
Using the concept of de Broglie wavelength under relativistic conditions to determine the spatial limits for electron range.
de Broglie wave-length [tex]\lambda = \dfrac{h}{p}[/tex]
where;
h = plank's constant
p = momentum of particle which is expressed as:
[tex]p = \dfrac{\sqrt{k^2+2kmc^2}}{c}[/tex]
replacing the expression for (p) into [tex]\lambda = \dfrac{h}{p}[/tex], we have:
[tex]\lambda = \dfrac{h}{\dfrac{\sqrt{k^2 +2kmc^2}}{c}}[/tex]
[tex]\lambda = \dfrac{hc}{\sqrt{k^2+2kmc^2}} --- (1)[/tex]
here;
c = velocity of light = 3 × 10⁸ m/s
h = 4.13 × 10⁻¹⁵ eV.s
i.e.
hc = (4.13 × 10⁻¹⁵ eV.s)(3 × 10⁸ m/s)
hc = 1240 eV.m
The electron's rest energy (mc²) = 0.511 × 10⁶ eV
For the elctrons;
the minimum accelerated voltage = 40kV
the maximum accelerated voltage = 100 kV
the minimum K.E of the electron K = eΔV
K = e × 40 kV
K = 40 KeV
K = 40 × 10³ eV
From equation (1);
[tex]\lambda = \dfrac{hc}{\sqrt{k^2 +2kmc^2}}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{\sqrt{(40 \times 10^3 \ eV)^2 +2(40 \times 10^3 \ eV)(0.511 \times 10^6 \ eV})}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{\sqrt{((1600000000) +(80000\times 511000))eV}}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{\sqrt{((1600000000) +(40880000000)eV}}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{\sqrt{(42480000000) \ eV}}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{206106.769 \ eV}}[/tex]
[tex]\lambda = 0.062 \ nm[/tex]
[tex]\mathbf{\lambda = 6.20 \ pm}[/tex]
The maximum K.E of the electron K = eΔV
K = e × 10 kV
K = 10 KeV
K = 100 × 10³ eV
From equation (1);
[tex]\lambda = \dfrac{hc}{\sqrt{k^2 +2kmc^2}}[/tex]
[tex]\lambda = \dfrac{1240 \ eV.nm}{\sqrt{(100 \times 10^3 \ eV)^2 +2(100 \times 10^3 \ eV)(0.511 \times 10^6 \ eV})}[/tex]
[tex]\lambda = 0.037 \ nm[/tex]
[tex]\mathbf{\lambda = 3.70 \ pm}[/tex]
As a result, the spatial limits for the electron's range span from 6.02 pm to 3.70 pm.
