A) \[\frac{{{B}^{2}}}{2V{{E}^{2}}}\]
B) \[\frac{2V{{B}^{2}}}{{{E}^{2}}}\]
C) \[\frac{2V{{E}^{2}}}{{{B}^{2}}}\]
D) \[\frac{{{E}^{2}}}{2V{{B}^{2}}}\]
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 \[=\frac{1}{2}m{{v}^{2}}.\] 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 \[\frac{e}{m}=\frac{{{E}^{2}}}{2V{{B}^{2}}}\] Note : - (where V is the potential difference between cathode and anode)
You need to login to perform this action.
You will be redirected in
3 sec
