Question

A particle carrying 5 electrons starts from rest and is accelerated through a potential difference of 8900V. Calculate the K.E. acquired by it in MeV.$(Change$ $on$ $electron=1.6\times {10}^{-19}C)$

Answer 1

Let us assume particle moved from A to B

Where $V_A$ - $V_B$ = $8900V$

{$V_A$ and $V_B$ is Potential at A and B Respectively}

$q$ = charge on 5 electrons = $5\times 1.6\times {10}^{-19}C$

[Also $1eV=1.60\times {10}^{-19}J$]

Therefore,

Potential energy at A ($P.E{.}_A$) = $V_A\times q$ $J$

Potential energy at B ($P.E{.}_B$) = $V_B\times q$ $J$

Kinetic Energy at A ($K.E{.}_A$) = 0 (Since, Particle starts from rest)

Kinetic Energy at B ($K.E{.}_B$) = E

Applying Conservation of Energy,

Total Energy at A = Total Energy at B

$K.E{.}_A$ + $P.E{.}_A$ = $K.E{.}_B$ + $P.E{.}_B$

$\to$ 0 + $V_A\times q$ = E + $V_B\times q$

$\to$ E = $V_A\times q$ - $V_B\times q$

$\to$ E = $q$ $\times$ ($V_A$ - $V_B$)

$\to$ E = $5\times 1.6\times {10}^{-19}C$ $\times$ $\left(8900V\right)$

$\to$ E = $71200\times {10}^{-19}J$

$\to$ E = $\frac{71200\times {10}^{-19}}{1.60\times {10}^{-19}}$ $eV$

$\to$ E = $44500$ $eV$

$\to$ E = $0.0445\times {10}^6$ $eV$

$\to$ E = $0.0445$ $MeV$