The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
The thickness of the aluminum can be determined using from distance of closest approach of the particle.
[tex]K.E = \frac{2KZe^2}{r}[/tex]
where;
[tex]r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m[/tex]
Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.
Charge of alpah particle = 2e
[tex]r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m[/tex]
Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
Learn more about closest distance of approach here: https://brainly.com/question/6426420
