Question

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of −3.45×10−3 V .
The charge and the mass of an alpha particle are qα = 3.20×10−19 C and mα = 6.68×10−27 kg , respectively.

If you had carried out the algebra using variables before plugging numbers into your expressions, you would have found that

(vf)α=−2qαΔVmα−−−−−−−√,

where ΔV is measured in volts. To verify that this expression for (vf)α has the correct units of velocity, you need to perform some unit analysis. Begin by finding the equivalent of a volt in terms of basic SI units. What is a volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (C)?

answer includes C

Answer 1

Answer:

v = 5.75 10² m / s

Explanation:

For this exercise let's use the concepts of energy conservation

initial with the particle at rest

    Em₀ = U = -q DV

final

    $Em_{f}$ = k = ½ m v²

    Em₀ =  $Em_{f}$

    q V = ½ m v²

    v = √(-2 qV /m)

    v = √ (-2  3.20 10⁻¹⁹ (-3.45 10⁻³) / 6.68 10⁻²⁷)

    v = 0.575 10³ m / s

    v = 5.75 10² m / s

Let's perform a dimensional analysis

    v = [m / s]

    q = [C]

    V = [N m / C]

    m = [kg]

    [m / s] = √ ([C] [N m / C] / kg) = √ ([N m / kg])

Newton's units are

    [N] = [kg m / s2]

we replace

    [m / s] = √ ([(kg m / s2) m / kg) = √ ([m2 / s2])

   [m / s] = [m / s]

what will confirm the units are correct