Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron (mass $m$) with initial velocity $\mathbf{v}=v_0\hat{\mathbf{i}}+v_0\hat{\mathbf{j}}$ is in an electric field $\mathbf{E}=-E_0\hat{\mathbf{k}}$. If ${\lambda }_0$ is initial de-Broglie wavelength of electron, then its de Broglie wavelength at time $t$ is given by

Imagine you are tracking an electron moving through space. Initially, it is cruising along the $xy$-plane with a velocity vector ${\mathbf{v}}_0=v_0\hat{\mathbf{i}}+v_0\hat{\mathbf{j}}$. This means it has equal speed components along both the $x$ and $y$ axes.

Suddenly, it enters a region with a uniform electric field pointing straight down along the negative $z$-axis, given by $\mathbf{E}=-E_0\hat{\mathbf{k}}$. Our goal is to find out how the electron's de Broglie wavelength changes over time due to this electric field.

The de Broglie wavelength $\lambda$ of a particle is fundamentally linked to its momentum. The relationship is elegantly simple:

Before the electric field starts affecting the electron, its initial speed is simply the magnitude of its initial velocity vector:

Therefore, the initial de Broglie wavelength, ${\lambda }_0$, is:

Now, let's see how the electric field alters the electron's journey. The electric field exerts a force on the electron. Because the electron carries a negative charge ($-e$), the force it experiences is in the exact opposite direction of the electric field:

This force gives the electron a constant acceleration strictly along the positive $z$-axis:

Notice something crucial here: there is absolutely no force acting in the $x$ or $y$ directions. This means the $x$ and $y$ components of the electron's velocity will remain perfectly constant at $v_0$. Only the $z$-component will change, growing linearly with time. Using the first equation of motion, $\mathbf{v}(t)={\mathbf{v}}_0+\mathbf{a}t$, we can write the velocity vector at any time $t$:

To find the new de Broglie wavelength, we need the new overall speed of the electron. We calculate the magnitude of this new velocity vector:

Finally, we substitute this new speed back into our de Broglie wavelength formula:

To make this look like our options and relate it back to ${\lambda }_0$, we need to factor out $2v_0^2$ from inside the square root:

Recognizing that the term $\frac{h}{m{v}_0\sqrt{2}}$ is exactly our initial wavelength ${\lambda }_0$, we arrive at our elegant final answer: