Answer:
C: Stop before entering the pedestrian crosswalk.
Explanation:
The answer is A. Immediately inform her colleague
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.
<h3>
Thickness of the aluminum</h3>
The thickness of the aluminum can be determined using from distance of closest approach of the particle.

where;
- Z is the atomic number of aluminium = 13
- e is charge
- r is distance of closest approach = thickness of aluminium
- k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

<h3>For 2.5 MeV protons</h3>
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.
<h3>For 10 MeV alpha-particles</h3>
Charge of alpah particle = 2e

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: brainly.com/question/6426420
(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
<u />
Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell