QJEEM22Ph001a: JEE Main 2022 Physics - Class 12 JEEM22Ph001a Solution: The equation can be used to find the de-Brogli wavelength of an …

The de Broglie wavelength of a particle can be expressed as:

$\lambda = \frac{h}{p}$

where:

• $h$ is Planck's constant, and
• $p$ is the momentum of the particle.

For an electron accelerated through a potential difference of $V$ volts, its kinetic energy $K$ is given by:

$K = eV$

where $e$ is the charge of an electron.

The electron's momentum can be expressed in terms of its kinetic energy as:

$p = \sqrt{2mK}$

where $m$ is the mass of the electron.

Substituting this into the de Broglie wavelength equation, we get:

$\lambda = \frac{h}{\sqrt{2mK}}$

Comparing this with the given equation $\lambda = \frac{1.227}{x} \, nm$ , we see that $x$ must correspond to:

$x = \sqrt{2mK}$

Since $K = eV$ , we can substitute this into our expression for $x$ to get:

$x = \sqrt{2meV} = \sqrt{V}$

where we've used the fact that the mass and charge of an electron are constants.