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Electrostatics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsElectrostatics3 Marks
Question
A proton is accelerated from rest through a potential difference of $500$ volts. Find its final momentum. $\left[ m _{ p }=1.67 \times 10^{-27} kg , e =1.6 \times 10^{-19} C \right]$
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Answer

$m _{ p }=1.67 \times 10^{-27} \ kg ,$
$e =1.6 \times 10^{-19} C ,$
$u =0,$
$V =500 V$
Initial $K E , KE _i=\frac{1}{2} m _{ p } u ^2=0$
$\therefore \Delta K E=K E_f-K E_{ i }= KE _{ f }$
$\Delta K E= qV$
$\therefore KE _{ f }= qV KE_{f } =\frac{1}{2} m _{ p } v ^2=\frac{p_{ f }^2}{2 m_{ p }}$
where $p _{ f }= m _{ p } v \equiv$ the magnitude of the final momentum of the proton.
$\therefore P _{ f }{ }^2=2 m _{ p } qV$
$\therefore P _{ f }=\sqrt{2 m_{ p } q V}$
$=\sqrt{2\left(1.67 \times 10^{-27} \ kg \right)\left(1.6 \times 10^{-19} C \right)(500 V )}$
$=5.169 \times 10^{-22} \ kg \cdot m / s$
The momentum is directed along the applied electric field.

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