Q. If the kinetic energy of a moving particle is $E$ , then the de-Broglie wavelength is
NTA AbhyasNTA Abhyas 2022
Solution:
Kinetic energy of the moving particle is given by, $E=\frac{1}{2}mv^{2}$
$\Rightarrow mv=\sqrt{2 m E}...\left(1\right)$
where, $mv$ is the momentum of a moving particle.
de-Broglie wavelength of a moving particle is given by, $\lambda =\frac{h}{m v}...\left(2\right)$
where, $h$ is a Planck's constant.
Using equation $\left(1\right)$ and $\left(2\right)$ ,
$\Rightarrow \lambda =\frac{h}{\sqrt{2 m E}}$
$\Rightarrow mv=\sqrt{2 m E}...\left(1\right)$
where, $mv$ is the momentum of a moving particle.
de-Broglie wavelength of a moving particle is given by, $\lambda =\frac{h}{m v}...\left(2\right)$
where, $h$ is a Planck's constant.
Using equation $\left(1\right)$ and $\left(2\right)$ ,
$\Rightarrow \lambda =\frac{h}{\sqrt{2 m E}}$