A proton with an initial speed of 900,000ms is brought to rest by an electric...

Step 1
Solution:
a) ${\displaystyle \frac{1}{2}{m}_p{v}^2=qV\Rightarrow V=\frac{m_p{v}^2}{zq}=\frac{1.67\times {10}^{-27}\times {(9\times {10}^5)}^2}{2\times 1.6\times {10}^{-19}}v}$
${\displaystyle \Rightarrow v=4227\times 19v}$
b) Initial kinetic energy of proton
${\displaystyle =4227\times 19eV}$

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