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 \[r=\frac{\sqrt{2mK}}{qB}\] where K = Kinetic energy of electron \[\Rightarrow \,\,K=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}\,\,=\,\,\left( \frac{e}{m} \right)\,\frac{e{{B}^{2}}{{r}^{2}}}{2}\] \[=\,\,\frac{1}{2}\times 1.7\times {{10}^{11}}\times 1.6\times {{10}^{-19}}\] \[\times \,{{\left( \frac{1}{\sqrt{17}}\times {{10}^{-5}} \right)}^{2}}\,\times \,{{(1)}^{2}}\] \[=\,\,8\times {{10}^{-}}^{20}\,J\,\,=\,\,0.5\,eV\] By using \[E={{W}_{0}}+{{K}_{\max }}\] \[\Rightarrow \,\,\,{{W}_{0}}=E-{{K}_{\max }}\] \[=\,\,\left( \frac{12375}{2475} \right)\,eV-0.5\,\,eV\,\,=\,\,4.5\,eV\,\]You need to login to perform this action.
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