1. Tardigrade
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  4. The equation λ=(1.227/ x ) nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for: Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V= Accelerating potential in volts for electron

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Q. The equation $\lambda=\frac{1.227}{ x } nm$ can be used to find the de-Brogli wavelength of an electron.
In this equation $x$ stands for :
Where,
$m =$ mass of electron
$P =$ momentum of electron
$K =$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

JEE MainJEE Main 2022Dual Nature of Radiation and Matter

Solution:

$ \left.\lambda=\frac{ h }{ m v} \text { (de-Broglie's wavelength }\right)$
$ \lambda \frac{ h }{\sqrt{2 m ( K \cdot E )}}$
$ h =\frac{ h }{\sqrt{2 mqV }}$
Putting the values of $m ; q$
We get $\lambda=\frac{1.22}{\sqrt{ V }} nm$