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

Question

The equation $\lambda=\frac{1.227}{x} \mathrm{~nm}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for :

Where

$\mathrm{m}=$ mass of electron

$\mathrm{P}=$ momentum of electron

$\mathrm{K}=$ Kinetic energy of electron

$\mathrm{V}=$ Accelerating potential in volts for electron

$\sqrt{\mathrm{mK}}$

$\sqrt{\mathrm{P}}$

$\sqrt{\mathrm{K}}$

$\sqrt{\mathrm{V}}$

Please login to view the detailed solution steps...