A) 3.42cm
B) 4.23cm
C) 5.17cm
D) 7.7cm
Correct Answer: A
Solution :
KE of electron \[KE=6\times {{10}^{-16}}J\] \[B=6\times {{10}^{-3}}Wb/{{m}^{2}}\] \[q=1.6\times {{10}^{-19}}C\] \[m=9\times {{10}^{-31}}kg\] Radius of circular path, \[r=\frac{mv}{Bq}\] But \[KE=\frac{1}{2}m{{v}^{2}}\] \[\therefore \] \[v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 6\times {{10}^{-16}}}{9\times {{10}^{-31}}}}\] \[=\sqrt{\frac{4}{3}\times {{10}^{15}}}\] \[=3.6\times {{10}^{7}}m/s\] \[\therefore \] \[r=\frac{9\times {{10}^{-31}}\times 3.6\times {{10}^{7}}}{6\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}}\] \[=3.42\times {{10}^{-2}}m\] \[=3.42cm\]
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