Question

The mass of a particle is 400 times than that of an electron and charge is double. The particle is accelerated by \(5 \mathrm{~V}\). Initially the particle remained at rest, then its final kinetic energy is \(\ldots \ldots \ldots\) (A) \(5 \mathrm{eV}\) (B) \(10 \mathrm{eV}\) (C) \(100 \mathrm{eV}\) (D) \(2000 \mathrm{eV}\)

Short answer

The final kinetic energy of the particle after being accelerated is \(10 eV\).

Step-by-step solution

Calculate the work done by the electric field

Given: Mass of the electron, \(m_e = 9.1 \times 10^{-31} kg\) Charge of the electron, \(e = 1.6 \times 10^{-19} C\) Mass of the particle, \(m_p = 400 m_e\) Charge of the particle, \(q_p = 2e = 3.2 \times 10^{-19} C\) The particle is accelerated by a potential difference of 5 V. The work done by the electric field on the charged particle is given by: \[W = q_pV\]

Use the work-energy theorem to find the final kinetic energy of the particle

According to the work-energy theorem, the work done on a body is equal to the change in its kinetic energy. Since the initial velocity of the particle is 0, the change in kinetic energy is equal to the final kinetic energy, \(K_f\). \[K_f = W\] Now, substitute all the given values to find the final kinetic energy: \[K_f = q_pV = (3.2 \times 10^{-19} C)(5 V) = 16 \times 10^{-19} J\] Convert this value to electron volts (eV) using the conversion factor \(1 eV = 1.6 \times 10^{-19} J\): \[K_f = \frac{16 \times 10^{-19} J}{1.6 \times 10^{-19} J/eV} = 10 eV\] So, the final kinetic energy of the particle is \(10 eV\). Therefore, the correct option is (B).