A) 200 keV
B) 50 keV
C) 100 keV
D) 25 keV
Correct Answer: C
Solution :
In this case magnetic force provides necessary centripetal force ie, \[qvB=\frac{m{{v}^{2}}}{T}\] Radius of path \[r=\frac{mv}{Bq}=\frac{\sqrt{2mE}}{qB}\] \[r=\frac{\sqrt{2mE}}{Bq}=\frac{\sqrt{2{{m}_{1}}{{E}_{1}}}}{Bq}\]or \[{{E}_{1}}=\frac{mE}{{{m}_{1}}}=\frac{(2{{m}_{1}})}{{{m}_{1}}}\times 50\,\text{keV}\,\] \[[\because \,m=2{{m}_{1}}]\]\[=100\,\text{keV}\]
You need to login to perform this action.
You will be redirected in
3 sec
