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Question: The de Broglie Wavelength of photon is twice the de Broglie wavelength of an electron. The speed of ...

The de Broglie Wavelength of photon is twice the de Broglie wavelength of an electron. The speed of the electron is Ve=c100.ThenV_{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(c/100)=100hmec.......(i)\lambda_{e} = \frac{h}{m_{e}v_{e}} = \frac{h}{m_{e}(c/100)} = \frac{100h}{m_{e}c}.......(i)

Ee=12meve2ormeve=2EemeE_{e} = \frac{1}{2}m_{e}{v_{e}}^{2}orm_{e}v_{e} = \sqrt{2E_{e}m_{e}}

λe=hmeve=h2meEeorEe=h22λc2me......(ii)\lambda_{e} = \frac{h}{m_{e}v_{e}} = \frac{h}{\sqrt{2m_{e}E_{e}}}orE_{e} = \frac{h^{2}}{2{\lambda_{c}}^{2}m_{e}}......(ii)

For photon

Ep=hcλp=hc2λe(∵λp=2λe(Given))E_{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 (using (i))

∴EeEp=1100=10−2\therefore\frac{E_{e}}{E_{p}} = \frac{1}{100} = 10^{- 2}

For electron pe=mever=m×c100p_{e} = m_{e}v_{er} = m \times \frac{c}{100}

∴pemec=1100=10−2\therefore\frac{p_{e}}{m_{e}c} = \frac{1}{100} = 1{0^{- 2}}^{}

Thus option (2) is correct