A) 4 eV
B) 2 eV
C) 8 eV
D) 6 eV
Correct Answer: C
Solution :
Let m be the mass and q the charge of the particle moving in a magnetic field B, then the energy is given by \[E=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}\] where r is radius of circular path. For proton \[{{E}_{P}}=\frac{{{e}^{2}}{{B}^{2}}{{r}^{2}}}{2m}\] ???. (i) For a-particle \[{{E}_{\alpha }}=\frac{{{(2e)}^{2}}{{B}^{2}}{{r}^{2}}}{2(4m)}\] ???.. (ii) From Eqs. (i) and (ii), we,get \[\frac{{{E}_{\alpha }}}{{{E}_{p}}}=1\] \[\Rightarrow \] \[{{E}_{\alpha }}={{E}_{p}}=8\,eV\] (given)You need to login to perform this action.
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