Animated Solution for Chemistry - Atomic Structure: The wavelength of electrons accelerated from rest through a potential difference of $40\text{ kV}$ is $x\times 10^{-12}\text{ m}$. The value of $x$ is ...... (Nearest integer) Given : Mass of electron $=9.1\times 10^{-31}\text{ kg}$ Charge on an electron $=1.6\times 10^{-19}\text{ C}$ Planck's constant $=6.63\times 10^{-34}\text{ Js}$

Have you ever wondered what happens when you take a tiny, seemingly insignificant electron and blast it with a massive $40,000$ volts of electricity? It doesn't just speed up; it undergoes a profound identity crisis. It stops acting purely like a little billiard ball and starts rippling through space like a wave. This is the mind-bending reality of wave-particle duality, and today, we are going to calculate exactly how that wave behaves.

Imagine an electron, initially at rest, placed in an electric field created by a massive $40\text{ kV}$ potential difference. As it accelerates, it gains kinetic energy. According to quantum mechanics, any moving particle has an associated matter wave, known as the de Broglie wave.

To find the wavelength of this matter wave, we use the foundational de Broglie equation:

Here, $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum of the electron. Since kinetic energy $K$ is related to momentum by $K=\frac{p^2}{2m}$, we can rewrite the momentum as $p=\sqrt{2mK}$. Substituting this back, we get:

But what is the kinetic energy here? The work done by the electric field on the electron is exactly equal to its charge $q$ times the accelerating potential $V$. By the work-energy theorem, this work is converted entirely into kinetic energy. So, we replace $K$ with $qV$:

This is our master formula. It beautifully connects the macroscopic world of voltage to the microscopic quantum world of wavelengths.

Now, let's carefully substitute the given values into our master formula. Don't get intimidated by the large exponents; we will handle them step-by-step.

Let's simplify the terms inside the square root first. Multiplying the numbers ($2\times 9.1\times 1.6\times 40$) gives $1164.8$. Adding the powers of ten ($-31-19+3$) gives $10^{-47}$. To make taking the square root easier, we can write this as $116.48\times 10^{-46}$.

Taking the square root, the denominator becomes approximately $10.79\times 10^{-23}$. Now, we just divide:

We can rewrite this in the format requested by the question:

According to the question, the wavelength is $x\times 10^{-12}\text{ m}$. Comparing this with our result, $x=6.14$. Rounding off to the nearest integer, we get $6$.

Here is a massive time-saver for JEE. For an electron accelerated through a potential $V$, you don't need to plug in all those constants every single time. You can directly use the derived shortcut formula:

Let's test it out. Substitute $40,000$ for $V$:

Since $1\text{ A˚}=10^{-10}\text{ m}$, we have:

The answer is still $6$, but you save a tremendous amount of calculation time! Always keep this formula in your arsenal.