Questions: A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? 2.2 x 10^5 m/s 3.1 x 10^5 m/s 9.6 x 10^5 m/s 1.1 x 10^5 m/s

Solution Steps

We need to find the speed gained by a proton when it is accelerated through an electric potential difference of 500 V. The options provided are different speeds, and we need to calculate the correct one.

The kinetic energy gained by the proton is equal to the electric potential energy it loses. The change in electric potential energy is given by:

\[ \Delta U = q \Delta V \]

where \( q \) is the charge of the proton (\(1.602 \times 10^{-19} \, \text{C}\)) and \( \Delta V \) is the potential difference (500 V).

The kinetic energy (\( K \)) gained by the proton is:

\[ K = \Delta U = q \Delta V = (1.602 \times 10^{-19} \, \text{C})(500 \, \text{V}) = 8.01 \times 10^{-17} \, \text{J} \]

The kinetic energy of the proton is also given by:

\[ K = \frac{1}{2} m v^2 \]

where \( m \) is the mass of the proton (\(1.673 \times 10^{-27} \, \text{kg}\)) and \( v \) is the speed. Solving for \( v \), we have:

\[ v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 8.01 \times 10^{-17} \, \text{J}}{1.673 \times 10^{-27} \, \text{kg}}} \]

Substitute the values into the equation:

\[ v = \sqrt{\frac{1.602 \times 10^{-16} \, \text{J}}{1.673 \times 10^{-27} \, \text{kg}}} = \sqrt{9.576 \times 10^{10} \, \text{m}^2/\text{s}^2} \]

\[ v \approx 3.1 \times 10^5 \, \text{m/s} \]

Final Answer