A) \[2\sqrt{3}\times {{10}^{5}}m{{s}^{-1}}\]
B) \[8\times {{10}^{5}}m{{s}^{-1}}\]
C) \[16\times {{10}^{5}}m{{s}^{-1}}\]
D) \[4\sqrt{2}\times {{10}^{5}}m{{s}^{-1}}\]
Correct Answer: D
Solution :
Given, \[{{m}_{\alpha }}=6.4\times {{10}^{-27}}kg,\] \[{{q}_{\alpha }}=3.2\times {{10}^{-19}}C,E=1.6\times {{10}^{5}}V{{m}^{-1}}\] Force on \[\text{ }\!\!\alpha\!\!\text{ -}\]particle \[{{q}_{\alpha }}=E=3.2\times {{10}^{-19}}\times 1.6\times {{10}^{5}}\] \[=51.2\times {{10}^{-15}}N\] Now, acceleration of the particle \[\alpha =\frac{F}{{{m}_{\alpha }}}=\frac{51.2\times {{10}^{-15}}}{6.4\times {{10}^{-27}}}\] \[=0.8\times {{10}^{13}}m{{s}^{-2}}\] \[\because \]Initial velocity, u = 0 \[\therefore \] \[{{v}^{2}}=2\alpha S\] \[=2\times 8\times {{10}^{12}}\times 2\times {{10}^{-2}}\] \[=32\times {{10}^{10}}\] or \[v=4\sqrt{2}\times {{10}^{5}}m{{s}^{-1}}\]
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