An alpha-particle of mass 6.4 times 10^{-27} | vedclass

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(D) Given: Mass of $\alpha$-particle $m = 6.4 	imes 10^{-27} \ kg$,charge $q = 3.2 	imes 10^{-19} \ C$,electric field $E = 1.6 	imes 10^{5} \ Vm^{-

Vedclass · Accepted Answer

$4 \sqrt{2} 	imes 10^{5} \ ms^{-1}$

Vedclass · Answer

(D) Given: Mass of $\alpha$-particle $m = 6.4 	imes 10^{-27} \ kg$,charge $q = 3.2 	imes 10^{-19} \ C$,electric field $E = 1.6 	imes 10^{5} \ Vm^{-1}$,and distance $s = 2 	imes 10^{-2} \ m$.The force acting on the $\alpha$-particle is $F = qE = (3.2 	imes 10^{-19}) 	imes (1.6 	imes 10^{5}) = 5.12 	imes 10^{-14} \ N$.The acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{5.12 	imes 10^{-14}}{6.4 	imes 10^{-27}} = 0.8 	imes 10^{13} \ ms^{-2} = 8 	imes 10^{12} \ ms^{-2}$.Using the equation of motion $v^2 = u^2 + 2as$,where initial velocity $u = 0$:$v^2 = 0 + 2 	imes (8 	imes 10^{12}) 	imes (2 	imes 10^{-2}) = 32 	imes 10^{10}$.Taking the square root,$v = \sqrt{32 	imes 10^{10}} = \sqrt{16 	imes 2} 	imes 10^{5} = 4 \sqrt{2} 	imes 10^{5} \ ms^{-1}$.

An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

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