A) Li+
B) Electron
C) Proton
D) \[H{{e}^{+}}\]
Correct Answer: A
Solution :
\[\nu =\frac{qB}{2\pi m}\Rightarrow \nu \propto \frac{q}{m}\] \[{{\left( \frac{q}{m} \right)}_{L{{i}^{+}}}}\]is minimum so \[={{B}_{1}}{{\left( 1+\frac{{{h}^{2}}}{{{r}^{2}}} \right)}^{-3/2}}={{B}_{1}}\left( 1-\frac{3}{2}.\frac{{{b}^{2}}}{{{r}^{2}}} \right)\] is minimum.
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