Questions: What potential difference is needed to give a helium nucleus (Q=3.2 × 10^-19 C) 132 keV of kinetic energy? Express your answer using two significant figures.

Solution Steps

The kinetic energy (\(K\)) of a charged particle can be related to the potential difference (\(V\)) it has been accelerated through by the equation: \[ K = QV \] where:

• \(K\) is the kinetic energy,
• \(Q\) is the charge of the particle,
• \(V\) is the potential difference.

Given the kinetic energy in kiloelectronvolts (keV), we need to convert it to joules (J). The conversion factor is: \[ 1 \, \text{eV} = 1.6022 \times 10^{-19} \, \text{J} \] \[ 132 \, \text{keV} = 132 \times 10^3 \, \text{eV} \] \[ K = 132 \times 10^3 \times 1.6022 \times 10^{-19} \, \text{J} \] \[ K = 2.1149 \times 10^{-14} \, \text{J} \]

Rearrange the equation \( K = QV \) to solve for \( V \): \[ V = \frac{K}{Q} \] Substitute the known values: \[ V = \frac{2.1149 \times 10^{-14} \, \text{J}}{3.2 \times 10^{-19} \, \text{C}} \] \[ V = 6.6094 \times 10^4 \, \text{V} \]

Round the potential difference to two significant figures: \[ V \approx 6.6 \times 10^4 \, \text{V} \]

Final Answer