# A proton with an initial speed of 600,000 m/s is brought to rest by an electric field. What was...

## Question:

A proton with an initial speed of 600,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the proton? What was the initial kinetic energy of the proton, in electron volts?

## Conservation of energy:

When a proton is moving with a constant velocity and it is brought to rest by an external electric field, then the initial kinetic energy of the proton must be equal to the electric potential energy of the proton due to the electric field.

We are given:

• The initial velocity of the proton is {eq}u\ =\ 600,000\ \rm m/s{/eq}
• The mass of the proton is {eq}m_p\ =\ 1.67\times 10^{-27}\ \rm kg{/eq}

Let,

The potential difference that stop the proton is {eq}\Delta V{/eq}

• The charge of the proton is {eq}q\ =\ +e\ =\ +1.6\times 10^{-19}\ \rm C{/eq}

Since the moving electron comes to rest due to the electric field, so from the law of conservation of energy, the kinetic energy of the proton must be converted into the electric potential energy of the proton due to the potential difference of the electric field.

Now, from conservation of energy:

\begin{align} q\Delta V\ &=\ \dfrac{1}{2}mu^2\\[0.3 cm] \Delta V\ &=\ \dfrac{mu^2}{2q}\\[0.3 cm] &=\ \dfrac{1.67\times 10^{-27}\ \rm kg\times (600,000\ \rm m/s)^2}{2\times 1.6\times 10^{-19}\ \rm C}\\[0.3 cm] &=\ \boxed{\color{green}{1.87875\times 10^3\ \rm V}} \end{align}

Now,

THe initial kinetic energy of the electron will be equal to the electric potential energy of the proton, so the kinetic energy of the proton is:

\begin{align} KE\ &=\ q\Delta V\\[0.3 cm] &=\ +e\times (1.87875\times 10^3\ \rm V)\\[0.3 cm] &=\ \boxed{\color{green}{+1.87875\times 10^3\ \rm eV}} \end{align} 