Question

An electron (mass $𝑚$) with initial velocity $\begin{array}{l}\overrightarrow{v}={\stackrel{}{v}}_0\hat{i}+{\stackrel{}{v}}_0\hat{j}\end{array}$ is in an electric field, $\begin{array}{l}\overrightarrow{E}=-E_0\hat{k}\end{array}$. If ${𝜆}_0$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time $t$ is given by

• A. $\begin{array}{l}\lambda =\frac{{\lambda }_0}{\sqrt{1+\frac{E_0^2{e}^2}{m^2{v}_0^2}{t}^2}}\end{array}$
• B. $\begin{array}{l}\lambda =\frac{{\lambda }_0\sqrt{2}}{\sqrt{1+\frac{E_0^2{e}^2}{m^2{v}_0^2}{t}^2}}\end{array}$
• C. $\begin{array}{l}\lambda =\frac{{\lambda }_0}{\sqrt{1+\frac{E_0^2{e}^2}{2m^2{v}_0^2}{t}^2}}\end{array}$
• D. $\begin{array}{l}\lambda =\frac{{\lambda }_0}{\sqrt{2+\frac{E_0^2{e}^2}{m^2{v}_0^2}{t}^2}}\end{array}$

Answer 1

The correct option is A

$\begin{array}{l}\lambda =\frac{{\lambda }_0}{\sqrt{1+\frac{E_0^2{e}^2}{m^2{v}_0^2}{t}^2}}\end{array}$

Step 1. Given data:

Initial velocity, $\begin{array}{l}\overrightarrow{v}={\stackrel{}{v}}_0\hat{i}+{\stackrel{}{v}}_0\hat{j}\end{array}$

Electric field, $\begin{array}{l}\overrightarrow{E}=-E_0\hat{k}\end{array}$

Step 2. The momentum of an electron:

$$\begin{gathered} 𝑝=𝑚𝑣 \\ \Rightarrow 𝑝=\frac{ℎ}{{𝜆}_0} \end{gathered}$$

$$\begin{gathered} \because V_i=\sqrt{{V_0}^2+{V_0}^2} \\ \Rightarrow V_i=\sqrt{2}{V}_0 \end{gathered}$$

Initially, $𝑚(\surd 2{𝑣}_0)=\frac{ℎ}{{𝜆}_0}$

STEP 3. Velocity as a function of time = ${𝑣}_0\widehat{𝑖}+{𝑣}_0\widehat{𝑗}+\left(\frac{𝑒{𝐸}_0}{𝑚}\right)𝑡\widehat{𝑘}$

Because only force is acting in $Z-di{r}^n$.

So velocity will be changed in $Z.di{r}^n$ and kept constant in $x-ydi{r}^n$.

$\therefore V_{net}=\sqrt{{V_0}^2+{V_0}^2+{\left(\frac{e{E}_0}{m}\right)}^2}$

$\Rightarrow 𝑝=\frac{ℎ}{{𝜆}_0}$

$\Rightarrow m{V}_{net}=\frac{h}{\lambda }$

Step 4. To calculate the wavelength:

$\Rightarrow \lambda =\frac{h}{m{V}_{net}}$

$\begin{array}{l}\Rightarrow \lambda =\frac{h}{m\sqrt{2v_0^2+\frac{E_0^2{e}^2}{m^2}{t}^2}}\end{array}$ $\left(\because {\lambda }_0=\frac{h}{m\sqrt{2}{V}_0}\right)$

$\begin{array}{l}\therefore \lambda =\frac{{\lambda }_0}{\sqrt{1+\frac{E_0^2{e}^2}{m^2{v}_0^2}{t}^2}}\end{array}$

Hence, the correct option is (A).