Dual Nature of Matter

JEE Main level, single correct, medium

30 of 200

Question 30 of 200

The kinetic energy of an electron, α−\alpha -particle and a proton are given as 4K,2K4K,2K and KK respectively. The de-Broglie wavelength associated with electron (λe),α−({\lambda }_{e}),\alpha -particle (λα)({\lambda }_{\alpha }) and the proton (λp)({\lambda }_{p}) are as follows:

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Answer: (B) λα<λp<λe{\lambda }_{\alpha }\lt {\lambda }_{p}\lt {\lambda }_{e}

The formula to calculate the de Broglie wavelength of a particle is given by
λ=h2mK...(1)\lambda =\dfrac{h}{\sqrt{2mK}}...\left(1\right)
Now, the masses of the given particles are interrelated by the following relations:
me≈mp1837mα=4mp\begin{matrix}{m}_{e} & \approx & \dfrac{{m}_{p}}{1837} \\ {m}_{\alpha } & = & 4{m}_{p}\end{matrix}
The de Broglie wavelength for electron is given by
λe=h2me(4K)=h81837mpK...(2)\begin{matrix}{\lambda }_{e} & = & \dfrac{h}{\sqrt{2{m}_{e}\left(4K\right)}} \\ & = & \dfrac{h}{\sqrt{\dfrac{8}{1837}{m}_{p}K}}...\left(2\right)\end{matrix}
The de Broglie wavelength for proton is given by
λp=h2mp(2K)=h4mpK...(3)\begin{matrix}{\lambda }_{p} & = & \dfrac{h}{\sqrt{2{m}_{p}\left(2K\right)}} \\ & = & \dfrac{h}{\sqrt{4{m}_{p}K}}...\left(3\right)\end{matrix}
And, the de Broglie wavelength of alpha particle is given by
λα=h2mαK=h2(4mp)K=h8mpK...(4)\begin{matrix}{\lambda }_{\alpha } & = & \dfrac{h}{\sqrt{2{m}_{\alpha }K}} \\ & = & \dfrac{h}{\sqrt{2\left(4{m}_{p}\right)K}} \\ & = & \dfrac{h}{\sqrt{8{m}_{p}K}}...\left(4\right)\end{matrix}
From equation (2), (3) and (4), it can be concluded that λe>λp>λα{\lambda }_{e}\gt {\lambda }_{p}\gt {\lambda }_{\alpha }.

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