A) \[\sqrt{2}R\]
B) \[\frac{R}{\sqrt{2}}\]
C) \[\frac{R}{2}\]
D) R
Correct Answer: B
Solution :
As charge on both proton and deuteron is same i.e. 'e' Energy acquired by both, E = eV For Deuteron. Kinetic energy, \[=\frac{1}{2}m{{V}^{2}}=eV\] [V is the potential difference]\[v=\sqrt{\frac{2eV}{{{m}_{d}}}}\] But\[{{m}_{d}}=2m\] Therefore, \[v=\sqrt{\frac{2eV}{2m}}=\sqrt{\frac{eV}{m}}\] Radius of path, \[R=\frac{mv}{eB}\] Substituting value of V we get\[R=\frac{2m\sqrt{\frac{ev}{m}}}{eB}\] \[\frac{R}{2}=\frac{m\sqrt{\frac{ev}{m}}}{eB}\] ?(i) For proton : \[\frac{1}{2}m{{V}^{2}}=eV\] \[V=\sqrt{\frac{2eV}{m}}\] Radius of path, \[R'=\frac{mV}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}\] \[R'=\sqrt{2}\times \frac{R}{2}\][From eq. (i)]\[R'=\frac{R}{2}\]You need to login to perform this action.
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