A) \[2.5\times {{10}^{-10}}\,N\]
B) \[8\times {{10}^{-11}}\,N\]
C) \[2.5\times 10{{~}^{-11}}\,N\]
D) \[8\times {{10}^{-~12}}\,N\]
Correct Answer: D
Solution :
\[{{F}_{m}}=qvB\] \[\frac{1}{2}m{{v}^{2}}\,=\,2MeV=2\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\,J\] \[\frac{1}{2}\times 1.6\times {{10}^{-27}}\,{{v}^{2}}=2\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\,\] \[{{v}^{2}}=4\times {{10}^{14}}\] \[v=2\times {{10}^{7}}\,\,m/sec\]
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