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Physics Question on de broglie hypothesis

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is ve=c100v_{e}=\frac{c}{100} then

A

EeEp=10−4\frac{E_{e}}{E_{p}}=10^{-4}

B

EeEp=10−2\frac{E_{e}}{E_{p}}=10^{-2}

C

pemec=10−1\frac{p_{e}}{m_{e} c}=10^{-1}

D

pemec=10−4\frac{p_{e}}{m_{e} c}=10^{-4}

Answer

EeEp=10−2\frac{E_{e}}{E_{p}}=10^{-2}

Explanation

Solution

For electron, λe=hmeve=hme(c100)=100hmec…(i)\lambda_{e}=\frac{h}{m_{e}v_{e}}=\frac{h}{m_{e}\left(c 100\right)}=\frac{100 h}{m_{e}c} \ldots\left(i\right) Ee=12meve2 or meve=2EemeE_{e}=\frac{1}{2}m_{e}v_{e}^{2}\, or\, m_{e}v_{e}=\sqrt{2E_{e}m_{e}} λe=hmeve=h2meEe\lambda_{e}=\frac{h}{m_{e} v_{e}}=\frac{h}{\sqrt{2m_{e}E_{e}}} or Ee=h22λe2me…(ii)E_{e}=\frac{h^{2}}{2\lambda_{e}^{2}m_{e}} \ldots\left(ii\right) For photon, Ep=hcλp=hc2λe(∵λp=2λeE_{p}=\frac{hc}{\lambda_{p}}=\frac{hc}{2\lambda_{e}} (\because\lambda_{p}=2\lambda_{e} (Given) ∴EpEe=hc2λe×2λe2meh2=λemech\therefore \frac{E_{p}}{E_{e}}=\frac{hc}{2\lambda_{e}}\times\frac{2\lambda_{e}^{2}m_{e}}{h^{2}}=\frac{\lambda_{e}m_{e}c}{h} =100hmec×mech=100=\frac{100h}{m_{e}c}\times\frac{m_{e}c}{h}=100 ∴EeEp=1100=10−2\therefore \frac{E_{e}}{E_{p}}=\frac{1}{100}=10^{-2} (Using (i)) For electron, pe=meve=me×c100p_{e}=m_{e}v_{e}=m_{e}\times\frac{c}{100} ∴pemec=1100=10−2\therefore \frac{p_{e}}{m_{e} c}=\frac{1}{100}=10^{-2} Thus, option (b)\left(b\right) is correct