A) 1.8 eV
B) 2.1 eV
C) 4.5 eV
D) 3.3 eV
Correct Answer: C
Solution :
Radius of circular path described by a charged particle in a magnetic field is given by \[r=\frac{\sqrt{2\,mK}}{qB}\] where \[\operatorname{K} =Kinetic energy of electron\] \[\Rightarrow \,\,\,\,\,\,\,K=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2\,m}=\left( \frac{e}{m} \right)\frac{e{{B}^{2}}{{r}^{2}}}{2}\] \[=\,\,\,\frac{1}{2}\times 1.7\times {{10}^{11}}\times 1.6\times {{10}^{-19}}\] \[\times \,{{\left( \frac{1}{\sqrt{17}}\times {{10}^{-5}} \right)}^{2}}\,\times \,\,{{(1)}^{2}}\] \[=\,\,\,8\times 10{{\,}^{-20}}\,J\,\,=\,\,0.5\,e\,V\] By using \[E={{W}_{0}}+{{K}_{\max }}\] \[\Rightarrow \,\,\,\,{{W}_{0}}=E-{{K}_{\max }}=\left( \frac{12375}{2475} \right)\,eV-0.5\,eV=4.5\,eV\]
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