Dual Nature of Matter

JEE Main level, single correct, medium

24 of 200

Question 24 of 200

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 2525% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

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Answer: (B) 18\dfrac{1}{8}

For photon,
EP=hcλp⇒λp=hcEP{E}_{P}=\dfrac{hc}{{\lambda }_{p}} \Rightarrow {\lambda }_{p}=\dfrac{hc}{{E}_{P}}
For electron
λe=hmeve=hve2Ke\begin{matrix}{\lambda }_{e} & = & \dfrac{h}{{m}_{e}{v}_{e}} \\ & = & \dfrac{h{v}_{e}}{2{K}_{e}}\end{matrix}
Given ve=0.25c{v}_{e}=0.25c. So,
λe=h×0.25c2Ke=hc8Ke\begin{matrix}{\lambda }_{e} & = & \dfrac{h\times 0.25c}{2{K}_{e}} \\ & = & \dfrac{hc}{8{K}_{e}}\end{matrix}
Also λp=λe{\lambda }_{p}={\lambda }_{e}. Thus,
hcEp=hc8Ke⇒KeEp=18\dfrac{hc}{{E}_{p}}=\dfrac{hc}{8{K}_{e}} \Rightarrow \dfrac{{K}_{e}}{{E}_{p}}=\dfrac{1}{8}

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