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NEET PHYSICSEasy

An electron of mass mm and charge ee is accelerated from rest through a potential difference VV in vacuum. The final speed of the electron will be:

A

V\sqrt{e/m}

B

\sqrt{eV/m}

C

\sqrt{2eV/m}

D

2eV/m

Step-by-Step Solution

When a charged particle is accelerated through a potential difference, the work done by the electric field appears as kinetic energy of the particle (Work-Energy Theorem) [Source 30, 73].

  1. Work Done: The work done (WW) on a charge ee moving through a potential difference VV is given by W=eVW = eV [Source 153].
  2. Kinetic Energy: This work is converted into the kinetic energy (KK) of the electron. Since it starts from rest, the final kinetic energy is K=12mv2K = \frac{1}{2}mv^2.
  3. Conservation of Energy: Equating work done to the gain in kinetic energy: eV=12mv2eV = \frac{1}{2}mv^2
  4. Solve for velocity (vv): v2=2eVmv^2 = \frac{2eV}{m} v=2eVmv = \sqrt{\frac{2eV}{m}} [Source 47].
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