Question

An electron of mass $m$ with initial velocity $\overrightarrow{V}={{V}_{0}}\widehat{i}\left( {{V}_{0}} > 0 \right)$ enters an electric field $\overrightarrow{E}=-{{E}_{0}}\widehat{i}\left( {{E}_{0}}=\text{constant} > 0 \right)$ at $t=0$. If ${{\lambda }_{0}}$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time $t$ is
$A)\text{ }{{\lambda }_{0}}\left( 1+\dfrac{e{{E}_{0}}}{m{{V}_{0}}}t \right)$
$B)\text{ }{{\lambda }_{0}}t$
$C)\text{ }\dfrac{{{\lambda }_{0}}}{\left( 1+\dfrac{e{{E}_{0}}}{m{{V}_{0}}}t \right)}$
$D)\text{ }{{\lambda }_{0}}$

Answer 1

Hint: This problem can be solved by finding out the force and the acceleration on the electron due to the electric field and using this acceleration, finding the velocity of the electron at a specific time, using which the de-Broglie wavelength can be found out.

Formula used:
$\overrightarrow{F}=q\overrightarrow{E}$
$\overrightarrow{F}=m\overrightarrow{a}$
$v=u+at$
$\lambda =\dfrac{h}{mv}$

Complete answer:
We will find the force and acceleration of the electron due to the electric field and from this find its velocity at a specific instant. Using this velocity, we can find the de Broglie wavelength of the electron at that instant.
The force $\overrightarrow{F}$ exerted by an electric field $\overrightarrow{E}$ on a charge $q$ is given by
$\overrightarrow{F}=q\overrightarrow{E}$ --(1)
Also, the force $\overrightarrow{F}$ on a body of mass $m$ is related to the acceleration $\overrightarrow{a}$ produced by the force as
$\overrightarrow{F}=m\overrightarrow{a}$ --(2)
For motion in one dimension, the velocity $v$ of a body that is subjected to a constant acceleration $a$ for a time $t$ is given by
$v=u+at$ --(3)
Where $u$ is the initial velocity of the body.
The de Broglie wavelength $\lambda $ of a body of mass $m$ and moving at a speed $v$ is given as
$\lambda =\dfrac{h}{mv}$ --(4)
Where $h=6.636\times {{10}^{-34}}J.s$ is the planck’s constant.
Now, let us analyze the question.
The mass of the electron is $m$.
The charge of the electron is $-e=-1.602\times {{10}^{-19}}C$.
The initial velocity of the electron is $\overrightarrow{V}={{V}_{0}}\widehat{i}$.
Therefore, the magnitude of the initial velocity is ${{V}_{0}}$.
The electric field is given by $\overrightarrow{E}=-{{E}_{0}}\widehat{i}$.
Let the time interval under consideration be $t$.
The initial de Broglie wavelength of the electron is ${{\lambda }_{0}}$.
Let the force on the electron due to the electric field be $\overrightarrow{F}$ and the corresponding acceleration produced be $\overrightarrow{a}$.
Let the velocity of the electron at time $t$ be $v$.
Let the de Broglie wavelength at time $t$ as $\lambda $.

Now, using (4), we get
${{\lambda }_{0}}=\dfrac{h}{m{{V}_{0}}}$ --(5)

Also, using (1), we get
$\overrightarrow{F}=-e\left( -{{E}_{0}}\widehat{i} \right)=e{{E}_{0}}\widehat{i}$ --(6)

Using (2), we get
$\overrightarrow{F}=m\overrightarrow{a}$ --(7)
Putting (6) in (7), we get
$e{{E}_{0}}\widehat{i}=m\overrightarrow{a}$
$\therefore \overrightarrow{a}=\dfrac{e{{E}_{0}}}{m}\widehat{i}$ --(8)
The magnitude of this acceleration will be
$\therefore a=\overrightarrow{\left| a \right|}=\left| \dfrac{e{{E}_{0}}}{m}\widehat{i} \right|=\dfrac{e{{E}_{0}}}{m}$ --(9)
Now, using (3), we get the speed $v$ at time $t$ as
$v={{V}_{0}}+\left( \dfrac{e{{E}_{0}}}{m} \right)t={{V}_{0}}+\dfrac{e{{E}_{0}}}{m}t$ --(10)
Again using (4), we get, the de Broglie wavelength $\lambda $ at time $t$ as
$\lambda =\dfrac{h}{mv}$ --(11)
Putting (10) in (11), we get
$\lambda =\dfrac{h}{m\left( {{V}_{0}}+\dfrac{e{{E}_{0}}}{m}t \right)}=\dfrac{h}{m{{V}_{0}}\left( 1+\dfrac{e{{E}_{0}}}{m{{V}_{0}}}t \right)}$ --(12)
Putting (5) in (12), we get
$\lambda =\dfrac{{{\lambda }_{0}}}{\left( 1+\dfrac{e{{E}_{0}}}{m{{V}_{0}}}t \right)}$
Therefore, we have got the required expression for the de Broglie wavelength of the electron.

So, the correct answer is “Option C”.

Note:
Students often make the mistake of writing the charge on the electron as $+e$ and not $-e$, that is, they do not take into consideration that the electron has negative charge and only consider the magnitude of the charge. This will give an entirely different expression for force on the electron due to the electric field, leading to wrong expressions for the acceleration, velocity and ultimately the de Broglie wavelength of the electron. Students must always be careful about writing the correct signs of the charges while solving problems including electrostatics and electrodynamics.