A) \[\frac{{{\lambda }_{0}}}{\sqrt{1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\]
B) \[\frac{{{\lambda }_{0}}\sqrt{2}}{\sqrt{1+\frac{{{e}^{2}}{{E}^{2}}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\]
C) \[\frac{{{\lambda }_{0}}}{\sqrt{1+\frac{{{e}^{2}}{{E}^{2}}{{t}^{2}}}{2{{m}^{2}}v_{0}^{2}}}}\]
D) \[\frac{{{\lambda }_{0}}}{\sqrt{2+\frac{{{e}^{2}}{{E}^{2}}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\]
Correct Answer: C
Solution :
\[v={{v}_{0}}\,\hat{i}+{{v}_{0}}\hat{j}+\frac{e{{E}_{0}}}{m}t\,\,\hat{k}\] \[\therefore \,\,\,\,\,\,\,\left| {\vec{v}} \right|=\sqrt{2v_{0}^{2}+\left( \frac{e{{E}_{0}}t}{m} \right)}\] Initially, \[{{\lambda }_{0}}=\frac{h}{m{{v}_{0}}\sqrt{2}}\] Now, \[\lambda =\frac{h}{m\sqrt{2v_{0}^{2}+\left( \frac{e{{E}_{0}}t}{m} \right)}}\] \[\frac{\lambda }{{{\lambda }_{0}}}=\frac{h}{m\sqrt{1+{{\left( \frac{e{{E}_{0}}t}{\sqrt{2}m{{v}_{0}}} \right)}^{2}}}}\] \[\lambda =\frac{{{h}_{0}}}{\sqrt{1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{2{{m}^{2}}v_{0}^{2}}}}\]
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