A) \[3.2\times {{10}^{-20}}J~\]
B) \[3.2\times {{10}^{-18}}J\]
C) \[3.2\times {{10}^{-19}}J\]
D) zero
Correct Answer: B
Solution :
The kinetic energy acquired by the particle is \[KE=q\Delta V\] where q is charge and \[\Delta V\] the change in potential difference. Given, \[q=e=1.6\times {{10}^{-19}}C\] \[\Delta V={{V}_{2}}-{{V}_{1}}=70-50=20V\] \[\therefore \] \[KE=1.6\times {{10}^{-19}}\times 20=3.2\times {{10}^{-18}}J\]
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