A) \[6.5\times {{10}^{15}}\,m/{{\sec }^{2}}\]
B) \[6.5\times {{10}^{13}}\,m/{{\sec }^{2}}\]
C) \[6.5\times {{10}^{11}}\,m/{{\sec }^{2}}\]
D) \[6.5\times {{10}^{9}}\,m/{{\sec }^{2}}\]
Correct Answer: A
Solution :
F = ma = qvB Þ \[a=\frac{qvB}{m}=\frac{1.6\times {{10}^{-19}}\times 2\times 3.4\times {{10}^{7}}}{1.67\times {{10}^{-27}}}\] = 6.5 ´ 1015m/sec2You need to login to perform this action.
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