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(B) The kinetic energy $K$ gained by a charged particle accelerated through a potential difference $V$ is given by the formula $K = qV$.Here,the charg

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$10$

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(B) The kinetic energy $K$ gained by a charged particle accelerated through a potential difference $V$ is given by the formula $K = qV$.Here,the charge of the particle is $q = 2e$,where $e$ is the elementary charge of an electron.The potential difference is $V = 5 \ V$.Substituting these values into the formula:$K = (2e) 	imes (5 \ V) = 10 \ eV$.Since the particle was initially at rest,its final kinetic energy is equal to the work done on it by the electric field,which is $10 \ eV$.

$A$ particle has a mass $400$ times that of an electron and a charge double that of an electron. It is accelerated by a potential difference of $5 \ V$. If the particle was initially at rest,what will be its final kinetic energy in $eV$?

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