Question

What potential difference is needed to give an alpha particle (composed of 2 protons and 2 neutrons) \(200 . \mathrm{keV}\) of kinetic energy?

Short answer

Answer: The potential difference needed to give an alpha particle 200 keV of kinetic energy is \(1 \times 10^5 \mathrm{V}\).

Step-by-step solution

Identify the relationship between potential difference and kinetic energy for charged particles

The potential energy (PE) of a charged particle in an electric field is given by: PE = qV, where q is the charge of the particle and V is the potential difference. When a charged particle accelerates in an electric field, the potential energy is transformed into kinetic energy (KE). Thus, the kinetic energy gained by the particle is equal to the potential energy calculated as: KE = qV. In the given problem, we are required to find the potential difference V needed to give the alpha particle a kinetic energy of 200 keV.

Find the charge of an alpha particle

An alpha particle consists of 2 protons and 2 neutrons. Neutrons have no charge, whereas the charge of each proton is \(+1.6 \times 10^{-19} \mathrm{C}\). Therefore, the total charge of an alpha particle (q) is: q = 2(proton charge) = \(2 \times 1.6 \times 10^{-19} \mathrm{C} = 3.2 \times 10^{-19} \mathrm{C}\).

Convert the kinetic energy into Joules

The kinetic energy given is in electron volts (eV), so we need to convert it to Joules (J). To do this, we can use the conversion factor: 1 eV = \(1.6 \times 10^{-19} \mathrm{J}\). Thus, the kinetic energy in Joules is: KE(J) = KE(eV) \(\times 1.6 \times 10^{-19} \mathrm{J}\). KE(J) = 200,000 eV \(\times 1.6 \times 10^{-19} \mathrm{J}\) = \(3.2 \times 10^{-14} \mathrm{J}\).

Calculate the potential difference (V)

Using the relationship KE = qV, we can now find the potential difference V: V = KE/q. V = \((3.2 \times 10^{-14} \mathrm{J})/(3.2 \times 10^{-19} \mathrm{C})\) = \(1 \times 10^5 \mathrm{V}\). The potential difference needed to give an alpha particle 200 keV of kinetic energy is \(1 \times 10^5 \mathrm{V}\).