A) \[{{\lambda }_{0}}\]
B) \[{{\lambda }_{0}}\sqrt{1+\frac{{{e}^{2}}\operatorname{E}_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}\]
C) \[\frac{{{\lambda }_{0}}}{\sqrt{1+\frac{{{e}^{2}}\operatorname{E}_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\]
D) \[\frac{{{\lambda }_{0}}}{\left( 1+\frac{{{e}^{2}}\operatorname{E}_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}} \right)}\]
Correct Answer: C
Solution :
| Option [c] is correct. |
| Explanation: According to the problem de-Broglie wavelength of electron at time t=0 is \[{{\lambda }_{0}}=\frac{h}{m{{v}_{0}}}\] |
| Electrostatic force on electron in electric field is \[{{\vec{F}}_{e}}=-e\vec{E}=-e{{E}_{0}}\hat{j}\] |
| The acceleration of electron, \[\vec{a}=\frac{{\vec{F}}}{m}=-\frac{e{{E}_{0}}}{m}\hat{j}\] It is acting along negative y-axis. |
| The initial velocity of electron along x-sods, \[{{v}_{{{x}_{0}}}}={{v}_{o}}\hat{i}\] |
| This component of velocity will remain constant as there is no force on electron in this direction. Now considering y-direction, initial velocity of electron along y-axis,\[{{v}_{{{y}_{0}}}}=0\] |
| Velocity of electron after time t along y-axis \[{{v}_{y}}=0+\left( -\frac{e{{E}_{0}}}{m}\operatorname{J} \right)t=-\frac{e{{E}_{0}}}{m}t\hat{j}\] |
| Magnitude of velocity electron after t is \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}={{\sqrt{v_{0}^{2}+\left( -\frac{e{{E}_{0}}}{m}t \right)}}^{2}}\] |
| \[\Rightarrow \] \[={{v}_{0}}\sqrt{1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}\] |
| de - Broglie wavelength, \[{{\lambda }_{0}}=\frac{h}{m{{v}_{0}}}\] |
| \[=\frac{h}{m{{v}_{0}}\sqrt{1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}=\frac{{{\lambda }_{0}}}{\sqrt{1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\] |
| \[\Rightarrow \] \[{\lambda }'=\frac{{{\lambda }_{0}}}{\left( 1+\frac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}} \right)}\] |
You need to login to perform this action.
You will be redirected in
3 sec
