JEE Main 2023 — Physics Modern Physics

The formula to calculate the de Broglie wavelength of a particle is given by

$$\lambda =\frac{h}{\sqrt{2mK}}...(1)$$λ=2mK​h​...(1)

Now, the masses of the given particles are interrelated by the following relations:

$$\begin{matrix}{m}_{e} & \approx & \frac{{m}_{p}}{1837} \\ {m}_{\alpha } & = & 4{m}_{p}\end{matrix}$$me​mα​​≈=​1837mp​​4mp​​

The de Broglie wavelength for electron is given by

$$\begin{matrix}{\lambda }_{e} & = & \frac{h}{\sqrt{2{m}_{e}(4K)}} \\ & = & \frac{h}{\sqrt{\frac{8}{1837}{m}_{p}K}}...(2)\end{matrix}$$λe​​==​2me​(4K)​h​18378​mp​K​h​...(2)​

The de Broglie wavelength for proton is given by

$$\begin{matrix}{\lambda }_{p} & = & \frac{h}{\sqrt{2{m}_{p}(2K)}} \\ & = & \frac{h}{\sqrt{4{m}_{p}K}}...(3)\end{matrix}$$λp​​==​2mp​(2K)​h​4mp​K​h​...(3)​

And, the de Broglie wavelength of alpha particle is given by

$$\begin{matrix}{\lambda }_{\alpha } & = & \frac{h}{\sqrt{2{m}_{\alpha }K}} \\ & = & \frac{h}{\sqrt{2(4{m}_{p})K}} \\ & = & \frac{h}{\sqrt{8{m}_{p}K}}...(4)\end{matrix}$$λα​​===​2mα​K​h​2(4mp​)K​h​8mp​K​h​...(4)​

From equation (2), (3) and (4), it can be concluded that ${\lambda }_{e}>{\lambda }_{p}>{\lambda }_{\alpha }$λe​>λp​>λα​.