Question

An electron and a proton are each accelerated from rest through a potential difference of 100 V. Afterward, which particle has the larger de Broglie wavelength? Explain.

Short answer

Electron will have a Higher de-broglie wavelength.

Step-by-step solution

Step 1. Given information is :An electron and a proton accelerated from rest through a potential difference of 100 V.

We need to find out the particle having the larger de Broglie wavelength.

Step 2. Using the formula of Wavelength and De-Broglie wavelength.

We know that :

$$\begin{gathered} \nu =\sqrt{\frac{2K}{m}} \\ and\lambda =\frac{h}{mv} \\ \Rightarrow \lambda =\frac{h}{m\sqrt{\frac{2K}{m}}}=\frac{h}{\sqrt{2mK}} \\ Asde-brogliewavelengthisinverselyproportionalto \\ massoftheparticle,Theparticlewithsmallermasswill \\ haveahigherwavelength \end{gathered}$$

As the mass of electron is smaller than that of proton, Electron will have a Higher de-broglie wavelength.