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) \[9\times {{10}^{22}}m/{{s}^{2}}\]
Correct Answer: B
Solution :
Acceleration or electron moving round the nucleus is \[a=\frac{{{v}^{2}}}{r}=\frac{{{(2.18\times {{10}^{6}})}^{2}}}{0.528\times {{10}^{-10}}}\] Given, \[v=2.18\times {{10}^{6}}m/s,\,r=0.528\times {{10}^{-10}}m\] Substituting the values in the relation, we have \[a=\frac{{{(2.18\times {{10}^{6}})}^{2}}}{0.528\times {{10}^{-10}}}\] \[\approx 9\times {{10}^{22}}m/{{s}^{2}}\]
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