A) \[1.5\times {{10}^{3}}m/s\]
B) \[3\times {{10}^{6}}m/s\]
C) \[6\times {{10}^{4}}m/s\]
D) \[2\times {{10}^{6}}m/s\]
Correct Answer: A
Solution :
Here, the magnetic force will provide the necessary centripetal force \[\therefore \] \[\left| \frac{\Delta p}{p} \right|=\left| \frac{\Delta \lambda }{\lambda } \right|\] Bqr = mv For electron and proton, the magnetic field B, charge q and radius r, all same. So, mv = constant i.e. \[\Rightarrow \] \[\frac{p'}{p}=\frac{0.20}{100}=\frac{1}{500}\] \[\Rightarrow \]You need to login to perform this action.
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