Walter Clyburn

2022-01-14

A proton with an initial speed of $900,000\frac{m}{s}$ is brought to rest by an electric field.
What was the potential difference that stopped the proton?
Express your answer to two significant figures and include the appropriate units.
What was the initial kinetic energy of the proton, in electron volts?
Express your answer using two significant figures.

Papilys3q

Expert

Step 1
Solution:
a) $\frac{1}{2}{m}_{p}{v}^{2}=qV⇒V=\frac{{m}_{p}{v}^{2}}{zq}=\frac{1.67×{10}^{-27}×{\left(9×{10}^{5}\right)}^{2}}{2×1.6×{10}^{-19}}v$
$⇒v=4227×19v$
b) Initial kinetic energy of proton
$=4227×19eV$

Serita Dewitt

Expert

Step 1
Initial speed of proton ${v}_{i}=900,000\frac{m}{s}$
final speedf problem ${v}_{f}=0$
Now,
${v}_{i}=900,00\frac{m}{s}$
${v}_{f}=0$
Kinetic energy
$K{E}_{i}=\frac{1}{2}{m}_{p}{v}_{i}^{2}$
$K{E}_{f}=\frac{1}{2}m{v}_{f}^{2}=0$
Potential Energy
$P{E}_{i}=0$
$P{E}_{f}=?$
Here, from Conservation of Energy. Kinetic Energy is Converted in to Potential Energy
$\therefore \frac{1}{2}{m}_{p}{v}_{i}^{2}=P{E}_{f}$
$\frac{1}{2}×1.67×{10}^{27}×{\left(900000\right)}^{2}=P{E}_{f}=677×{10}^{-18}$
$Tunles\left\{\begin{array}{l}kg\\ \left(m{l}_{s}{\right)}^{2}\end{array}$
Now, Potential difference $\mathrm{\Delta }v=\frac{\mathrm{\Delta }PE}{q}$
$=\frac{P{E}_{f}-P{E}_{i}}{q}$
Step 2
Potential difference to stop the proton given by
$\mathrm{\Delta }v=\frac{P{E}_{f}}{q}$
$\left(\because P{E}_{i}=0\right)$
$=\frac{{677}^{-41}×{10}^{-18}Ton\le }{1.6×{10}^{-19}c}$
$\mathrm{\Delta }v=4228v$
b) Now, initial kinetic energy $K{E}_{i}=\frac{1}{2}{m}_{p}{v}_{i}^{2}$
$K{E}_{i}=677.4×{10}^{-18}T$
$\le v=1.6×{10}^{-19}T$
$\therefore$ Kinetic energy is in ev is 4228 ev
i.e. 4228 ev

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