What is the speed of a proton that has been accelerated through $2300\mathrm{V}$?

• A. $1.2\times {10}^5\mathrm{m}/\mathrm{s}$
• B. $5.4\times {10}^5\mathrm{m}/\mathrm{s}$
• C. $7.6\times {10}^5\mathrm{m}/\mathrm{s}$
• D. $9.3\times {10}^5\mathrm{m}/\mathrm{s}$
• E. $15\times {10}^5\mathrm{m}/\mathrm{s}$
• F. $6.6\times {10}^5\mathrm{m}/\mathrm{s}$

Answer: (F)

Verified step by step guidance

1. Understand that the problem involves a proton being accelerated through an electric potential difference of 2300 V. This means the proton gains kinetic energy equal to the electric potential energy it loses.
2. Use the formula for kinetic energy gained by the proton: \( KE = qV \), where \( q \) is the charge of the proton (approximately \( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference (2300 V).
3. The kinetic energy \( KE \) can also be expressed as \( \frac{1}{2}mv^2 \), where \( m \) is the mass of the proton (approximately \( 1.67 \times 10^{-27} \) kg) and \( v \) is the speed of the proton.
4. Set the two expressions for kinetic energy equal to each other: \( qV = \frac{1}{2}mv^2 \). Solve for \( v \) by rearranging the equation: \( v = \sqrt{\frac{2qV}{m}} \).
5. Substitute the known values for \( q \), \( V \), and \( m \) into the equation to find the speed \( v \) of the proton. This will give you the speed in meters per second.