Question

A particle having charge equal to twice the charge on proton and mass equal to $7350$ times that of the electron is accelerated through a potential difference of $100$ V. Then, the increase in kinetic energy will be :

• A. $100$ eV
• B. $200$ eV
• C. $7.35$ keV
• D. $7.35$ $\times$ 10${}^5$ eV

Answer 1

Answer: (B)

$200$ eV

The particle is accelerated through a potential difference $V=100$ volts.

Charge of the particle $q=2e$

where $e$ is the charge of an electron.

Using conservation of energy, the increase in kinetic energy will be equal to the potential energy. i.e, $KE=qV=2e\times 100volt=200$ eV.