A) \[{{r}_{d}}={{r}_{P}}\sqrt{2}\]
B) \[{{r}_{d}}\frac{{{r}_{P}}}{\sqrt{2}}\]
C) \[{{r}_{d}}={{r}_{P}}\]
D) \[{{r}_{d}}=2{{r}_{P}}\]
Correct Answer: A
Solution :
Radius of circular path \[r=\frac{mv}{qB}\] ??(i) and kinetic energy \[KE=\frac{1}{2}m{{v}^{2}}\] or \[v=\sqrt{\frac{2KE}{m}}\] putting the value of v in Eq. (i) \[r=\frac{m}{qB}\sqrt{\frac{2KE}{m}}=\frac{\sqrt{2m\,KE}}{qB}\] ?..(ii) For proton \[{{(}_{1}}{{H}^{1}})\] and deuteron \[{{(}_{1}}{{H}^{2}})\] electronic charge are same and both have same kinetic energy. \[\therefore \] By Eq. (ii) \[r\propto \sqrt{m}\] or \[\frac{{{r}_{d}}}{{{r}_{p}}}=\sqrt{\frac{{{m}_{d}}}{{{m}_{p}}}}\] \[\because \] \[{{m}_{d}}=2{{m}_{p}}\] \[\therefore \] \[\frac{{{r}_{d}}}{{{r}_{p}}}=\sqrt{\frac{2{{m}_{p}}}{{{m}_{p}}}}\] \[\Rightarrow \] \[{{r}_{d}}=\sqrt{2}{{r}_{p}}\]
You need to login to perform this action.
You will be redirected in
3 sec
