A) \[3\sqrt{3}\times {{10}^{5}}m{{s}^{-1}}\]
B) \[5\times {{10}^{5}}m{{s}^{-1}}\]
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\[\alpha -\]particle \[F={{q}_{\alpha }}E=3.2\times {{10}^{-19}}\times 1.6\times {{10}^{5}}\] \[=51.2\times {{10}^{-15}}N\] 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,\[u=0\] \[\therefore \] \[{{v}^{2}}=2\alpha S\] \[=2\times 8\times {{10}^{12}}\times 2\times {{10}^{-12}}\] \[=32\times {{10}^{10}}\] or \[v=4\sqrt{2}\times {{10}^{5}}\,\,m{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
