What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?

  • A
    $\lambda_{e} = \lambda_{p} > \lambda_{\alpha}$
  • B
    $\lambda_{e} = \lambda_{p} = \lambda_{\alpha}$
  • C
    $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
  • D
    $\lambda_{e} < \lambda_{p} < \lambda_{\alpha}$

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The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$). (in $Å$)

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