A) 12cm
B) 16cm
C) 11cm
D) 18cm
Correct Answer: C
Solution :
[c] K.E. of electron=\[10\,eV\Rightarrow \frac{1}{2}m{{v}^{2}}=10\,eV\] \[\Rightarrow \,\,\frac{1}{2}(9.1\times {{10}^{-31}}){{v}^{2}}=10\times 1.6\times {{10}^{-19}}\] \[\Rightarrow \,\,{{v}^{2}}=\frac{2\times 10\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}\] \[\Rightarrow \,{{v}^{2}}=3.52\times {{10}^{12}}\Rightarrow v=1.88\times {{10}^{6}}m\] Also, we know that for circular motion \[\frac{m{{v}^{2}}}{r}=Bev\Rightarrow r=\frac{mv}{Be}=11\,cm\]
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