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(C) The kinetic energy $(K.E.)$ gained by a charged particle accelerated through a potential difference $(\Delta V)$ is given by the formula: $K.E. =

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The kinetic energy $(K.E.)$ gained by a charged particle accelerated through a potential difference $(\Delta V)$ is given by the formula: $K.E. = q \cdot \Delta V$. For an $\alpha$ particle, the charge is $q = +2e$. The given potential difference is $\Delta V = 10^6 \ V$. Substituting these values into the formula: $K.E. = (2e) 	imes (10^6 \ V) = 2 	imes 10^6 \ eV$. Since $10^6 \ eV = 1 \ MeV$, we have: $K.E. = 2 \ MeV$.

An $\alpha$ particle is accelerated through a potential difference of $10^6 \ V$. Then the kinetic energy $(K.E.)$ of the particle will be .......... $MeV$.

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