A) \[\frac{{{\lambda }_{0}}}{\left[ 1+\frac{e{{\operatorname{E}}_{0}}t}{m\ {{v}_{0}}} \right]}\]
B) \[{{\lambda }_{0}}\left[ 1+\frac{e{{\operatorname{E}}_{0}}t}{m{{v}_{0}}} \right]\]
C) \[{{\lambda }_{0}}\]
D) \[{{\lambda }_{0}}t\]
Correct Answer: A
Solution :
| Option [a] is correct. |
| Explanation: The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de-Broglie wave is given by |
| \[\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2m\operatorname{E}}}\] |
| As initial velocity of the electron is\[{{v}_{0}}\overset{}{\mathop{i}}\,\], the initial de-Broglie wavelength of electron, |
| \[\lambda =\frac{h}{m{{v}_{0}}}\] .(i) |
| Electrostatic force on electron in electric field is, |
| \[{{\vec{F}}_{e}}=-e\vec{E}=-e\left[ -{{E}_{0}}\overset{}{\mathop{i}}\, \right]=eE{{z}_{0}}\overset{}{\mathop{i}}\,\] |
| Acceleration of electron, \[\vec{a}=\frac{{\vec{F}}}{m}=\frac{e{{E}_{0}}\overset{}{\mathop{j}}\,}{m}\] |
| \[{{\vec{v}}_{e}}={{v}_{0}}\overset{}{\mathop{i}}\,+\left( \frac{e{{E}_{0}}\overset{}{\mathop{\operatorname{i}}}\,}{m} \right)t=\left( {{v}_{0}}+\frac{e{{E}_{0}}}{m}t \right)\overset{}{\mathop{i}}\,\] |
| \[\Rightarrow \] \[\vec{v}={{v}_{0}}\left( 1+\frac{e{{E}_{0}}}{m{{v}_{0}}} \right)\overset{}{\mathop{i}}\,\] |
| de - Broglie wavelength associated with electron at time t is \[\lambda =\frac{h}{mv}\] |
| \[\lambda =\frac{h}{m\left[ {{v}_{0}}\left( 1+\frac{e{{E}_{0}}}{m{{v}_{0}}}t \right) \right]}=\frac{\frac{h}{m{{v}_{0}}}}{\left( 1+\frac{e{{E}_{0}}}{m{{v}_{0}}}t \right)}\] |
| \[\lambda =\frac{{{\lambda }_{0}}}{\left[ 1+\frac{e{{E}_{0}}}{m{{v}_{0}}}t \right]}\operatorname{As}{{\lambda }_{0}}=\frac{h}{m{{v}_{0}}}\] |
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