Respuesta :
Answer:
16.7 x 10²V
Explanation:
The work-energy theorem states that the change in kinetic energy of a particle, Δ[tex]K_{E}[/tex], results in work done by the particle, W. i.e;
Δ[tex]K_{E}[/tex] = W ------------------(i)
But:
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]mv² -
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]m [v² - u²] ------------------(ii)
Where;
m = mass of particle (proton in this case) = 1.67 x 10⁻²⁷kg
v = final velocity of the particle = 6 x 10⁵m/s
u = initial velocity of the particle = 2 x 10⁵m/s
Substitute these values into equation (ii) as follows;
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(6 x 10⁵)² - (2 x 10⁵)²]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 x 10¹⁰) - (4 x 10¹⁰)]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 - 4) x 10¹⁰]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [32 x 10¹⁰]
Δ[tex]K_{E}[/tex] = (1.67 x 10⁻²⁷) [16 x 10¹⁰]
Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷J
Also:
W = qΔV ----------------(iii)
Where;
q = charge of the particle (proton) = 1.6 x 10⁻¹⁹C
ΔV = change in potential difference of the particle = V (from the question)
Substitute these values into equation (iii) as follows;
W = 1.6 x 10⁻¹⁹ x V
W = 1.6 x 10⁻¹⁹V -------------------(iv)
Now:
Substitute the values of Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷ and W in equation(iv) into equation (i)
26.72 x 10⁻¹⁷ = 1.6 x 10⁻¹⁹V
Solve for V;
V = (26.72 x 10⁻¹⁷) / (1.6 x 10⁻¹⁹)
V = 16.7 x 10²V
Therefore, the potential difference through which the proton fell was 16.7 x 10²V
The potential contrast between different locations represents the work or energy dissipated in the transmission of the unit amount of voltage from one point to another.
Following are the calculation of the potential difference:
Change in energy [tex], e^{v}=\frac{1}{2} \ m(v_2^2-v_1^2)[/tex]
[tex]1.602 \times ^{-19}=\frac{1}{2} \times 1.67 \times 10^{-27} (6^2-2^2)\times 10^{10}\\\\[/tex]
[tex]\to v \times 1.602 \times 10^{-19}=0.835 \times 10^{-17}\times 32\\\\[/tex]
[tex]\to v = \frac{0.835 \times 10^{-17}\times 32}{1.602 \times 10^{-19}}\\\\[/tex]
[tex]= \frac{0.835 \times 10^{2}\times 32}{1.602 }\\\\ = \frac{26.72 \times 10^{2} }{1.602 }\\\\=16.67\times 10^{2}[/tex]
Therefore the final answer is "[tex]\bold{16.67 \times 10^2}[/tex]".
Learn more:
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