Question

An electron (mass m) with an initial velocity \[\bar{v}={{v}_{0}}\hat{i}\] is in an electric field \[\bar{E}={{E}_{0}}\hat{j}\]. If \[{{\lambda }_{0}}=\dfrac{h}{m{{v}_{0}}}\] Its de Broglie wavelength at time \[t\] is given by,
A \[{{\lambda }_{0}}\]
B \[{{\lambda }_{0}}\sqrt{1+\dfrac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}\]
C \[\dfrac{{{\lambda }_{0}}}{\sqrt{1+\dfrac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}}}}\]
D \[\dfrac{{{\lambda }_{0}}}{\left( 1+\dfrac{{{e}^{2}}E_{0}^{2}{{t}^{2}}}{{{m}^{2}}v_{0}^{2}} \right)}\]

Answer 1

Hint: Based on the dual nature of matter, electrons have both the particle nature and wave nature. Hence we can use the equation of motion to get the velocity of an electron. Also we have the de Broglie equation for finding wavelength of a particle in terms of velocity. By substituting the electron velocity in the de Broglie equation we can find the wavelength.

Formula used:
\[\lambda =\dfrac{h}{mv}\]
\[{{v}_{2}}={{v}_{1}}+at\]
Equation for finding magnitude of a vector,
\[\left| {\vec{V}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]

Complete answer:
Given that,
Mass of electron \[=m\]
Initial velocity, \[\vec{v}={{v}_{0}}\hat{i}\]
Electric field,\[\vec{E}={{E}_{0}}\hat{j}\]
\[{{\lambda }_{0}}=\dfrac{h}{m{{v}_{0}}}\]
de Broglie wavelength formula is expressed as;
\[\lambda =\dfrac{h}{mv}\] ---------- 1
Where,
\[\text{ }\!\!\lambda\!\!\text{ - wavelength in meters}\]
\[\text{m - mass of the particle}\]
\[\text{v- velocity of the particle}\]
\[\text{h- Plank }\!\!'\!\!\text{ s constant = }6.62607\text{ }\times{ }{{10}^{-34}}~J\text{ }s\]
To find velocity, we have the equation of motion,
\[{{v}_{2}}={{v}_{1}}+at\] --------- 2
\[{{\text{v}}_{\text{2}}}\text{= final velocity}\]
\[{{\text{v}}_{\text{1}}}\text{= initial velocity}\]
\[\text{a = acceleration}\]
\[\text{t = time}\]
Here,
\[{{v}_{1}}=\vec{v}={{v}_{0}}\hat{i}\]
Let’s find the acceleration of an electron.
We have,
\[F=ma\]
Then,
Acceleration of electron, \[a=\dfrac{F}{m}\]------- 3
Force applied on electron by electric field \[F=eE\]
Where,\[e\] is the charge of an electron and \[E\] is the electric field.
Given, \[\vec{E}={{E}_{0}}\hat{j}\]
Then,
\[F=e{{E}_{0}}\hat{j}\]
Substitute \[F=e{{E}_{0}}\hat{j}\]in equation 3. We get,
\[a=\dfrac{e{{E}_{0}}}{m}\hat{j}\]----- 4
Substituting 4 in equation 2 we get,
\[{{\vec{v}}_{2}}={{v}_{0}}\hat{i}+\left( \dfrac{e{{E}_{0}}}{m}t \right)\hat{j}\]
Now find the magnitude of \[{{\vec{v}}_{2}}\].
\[\left| {{v}_{2}} \right|=\sqrt{\left( {{\left( {{v}_{0}} \right)}^{2}}+{{\left( \dfrac{e{{E}_{0}}}{m}t \right)}^{2}} \right)}\]-------- 5
Substituting 5 in equation 1, we get,
\[\lambda =\dfrac{h}{m\left| {{v}_{2}} \right|}\]
\[\lambda =\dfrac{h}{m\sqrt{\left( {{\left( {{v}_{0}} \right)}^{2}}+{{\left( \dfrac{e{{E}_{0}}}{m}t \right)}^{2}} \right)}}\]
\[\lambda =\dfrac{h}{m\sqrt{\left( {{\left( {{v}_{0}} \right)}^{2}}+\dfrac{{{e}^{2}}{{E}_{0}}^{2}{{t}^{2}}}{{{m}^{2}}} \right)}}\]
\[\lambda =\dfrac{h}{m{{v}_{0}}\sqrt{\left( 1+\dfrac{{{e}^{2}}{{E}_{0}}^{2}{{t}^{2}}}{{{m}^{2}}{{v}_{0}}^{2}} \right)}}\]
Here, it is given that, \[{{\lambda }_{0}}=\dfrac{h}{m{{v}_{0}}}\]
Then,
\[\lambda =\dfrac{{{\lambda }_{0}}}{\sqrt{\left( 1+\dfrac{{{e}^{2}}{{E}_{0}}^{2}{{t}^{2}}}{{{m}^{2}}{{v}_{0}}^{2}} \right)}}\]

So, the correct answer is “Option C”.

Note:
Analyzing the de Broglie relation \[\left( \text{ }\!\!\lambda\!\!\text{ =}\dfrac{\text{h}}{\text{p}}\text{ }\left( \text{Where p is the momentum} \right) \right)\], we can see that slow moving electrons have a large wavelength, and fast moving electrons have a short wavelength. This de Broglie relation can be applied only to smaller objects, as the wave properties are only observable in them.