A) \[\frac{{{B}^{2}}}{2V{{E}^{2}}}\]
B) \[\frac{2V{{B}^{2}}}{{{E}^{2}}}\]
C) \[\frac{2V{{E}^{2}}}{{{B}^{2}}}\]
D) \[\frac{2{{E}^{2}}}{2V{{B}^{2}}}\]
Correct Answer: D
Solution :
As the electron beam is not deflected, then \[{{F}_{m}}={{F}_{e}}\] or \[Bev=Ee\] or \[v=\frac{E}{B}\] ?(i) As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode = eV. Also, kinetic energy of the electron at anode \[v=\frac{1}{2}m{{v}^{2}}.\]According to law of conservation of energy \[\frac{1}{2}m{{v}^{2}}=eV\] or \[v=\sqrt{\frac{2eV}{m}}\] ?(ii) From Eqs. (i) and (ii), we have \[\sqrt{\frac{2eV}{m}}=\frac{E}{B}\] or \[\frac{e}{m}=\frac{{{E}_{2}}}{2V{{B}^{2}}}\]
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