Answer:
\[{{v}_{1}}\propto \left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\] or \[{{v}_{1}}\propto \frac{3}{4}\] (ii) \[{{v}_{2}}\propto \left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]\] or \[{{v}_{2}}\propto \frac{1}{4}\] \[\therefore \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{v}_{1}}}{{{v}_{2}}}\] \[=\frac{3}{4}\times \frac{4}{1}=3:1.\]
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