A) \[9\times {{10}^{18}}m/{{s}^{2}}\]
B) \[9\times {{10}^{22}}m/{{s}^{2}}\]
C) \[9\times {{10}^{-22}}m/{{s}^{2}}\]
D) \[9\times {{10}^{12}}m/{{s}^{2}}\]
Correct Answer: B
Solution :
Acceleration of the electron \[a=\frac{{{v}^{2}}}{r}\] \[v=2.18\times {{10}^{6}}m/s\] \[r=0.528\,\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=0.528\times {{10}^{-10}}m\] \[\therefore \] \[a=\frac{{{(2.18\,\times \,{{10}^{6}})}^{2}}}{0.528\times {{20}^{-10}}}\] \[=9\times {{10}^{22}}\,m/{{s}^{2}}\]
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