Dual Nature of Matter

JEE Main level, single correct, medium

67 of 200

Question 67 of 200

An electron (mass mm ) with initial velocity v⃗=v0i^+v0j^\vec{v}={v}_{0}\hat{i}+{v}_{0}\hat{j} is in an electric filed E⃗=−E0k^.\vec{E}=-{E}_{0}\hat{k}. If λ0{\lambda }_{0} is initial de-Broglie wavelength of electron, its de-Broglie wave length at time tt is given by:

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Answer: (C) λ01+e2E2t22m2v02\dfrac{{\lambda }_{0}}{\sqrt{1+\dfrac{{e}^{2}{E}^{2}{t}^{2}}{2{m}^{2}{v}_{0}^{2}}}}

Initial velocity of electron,
v=v0i^+v0j^⇒∣u∣=v02+v02=v02v={v}_{0}\hat{i}+{v}_{0}\hat{j}\Rightarrow \left|u\right|=\sqrt{{v}_{0}^{2}+{v}_{0}^{2}}={v}_{0}\sqrt{2}
Initial de Broglie wavelength associated with electron,
λ0=hp=hmv02{\lambda }_{0}=\dfrac{h}{p}=\dfrac{h}{m{v}_{0}\sqrt{2}}
now,
v=u+atv=u+at
Velocity as a function of time =v0i^+v0j^+eE0mtk^={v}_{0}\hat{i}+{v}_{0}\hat{j}+\dfrac{e{E}_{0}}{m}t\hat{k}
so wavelength λ=hm2v02+e2E02m2t2\lambda =\dfrac{h}{m\sqrt{2{v}_{0}^{2}+\dfrac{{e}^{2}{E}_{0}^{2}}{{m}^{2}}{t}^{2}}}
λ=λ01+e2E022m2v02t2\lambda =\dfrac{{\lambda }_{0}}{\sqrt{1+\dfrac{{e}^{2}{E}_{0}^{2}}{2{m}^{2}{v}_{0}^{2}}{t}^{2}}}

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