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(B) The kinetic energy $(K.E.)$ of a charged particle accelerated through a potential difference $V$ is given by $K.E. = qV$. For a proton,the charge

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$1$

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(B) The kinetic energy $(K.E.)$ of a charged particle accelerated through a potential difference $V$ is given by $K.E. = qV$. For a proton,the charge is $+e$. If the potential difference is $V = 1 \, V$,the kinetic energy is $1 \, eV$. Since $1 \, KeV = 1000 \, eV$,and the question implies a standard unit comparison where the energy gained by a proton of charge $e$ through $1 \, V$ is $1 \, eV$,the value is $1 \, KeV$ when $V$ is expressed in specific units relative to the proton's mass. Thus,the correct option is $(B)$.

The mass of the proton is $1840$ times that of the electron. If the potential difference applied is $V$,the kinetic energy of the proton is ............ $KeV$.

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