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Photon's de Broglie wavelength is twice the electron's (electron speed c/100): find Ep/Ee

The same question, asked 2 times: May 2021 S1, 26 Apr 2025 S2. Every wording below, oldest first.

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  1. Question 1

    May 2021 S1

    29
    Let EeE_e and EpE_p represent kinetic energy of electron and photon, respectively. If de Broglie wavelength λp\lambda_p of a photon is twice the de Broglie wavelength λe\lambda_e of an electron, then EpEe\frac{E_p}{E_e} is (speed of electron =c100= \frac{c}{100}, cc = velocity of light)

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  2. Question 2

    26 Apr 2025 S2

    15
    Let EeE_{e} and EpE_{p} represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then Ep/EeE_{p}/E_{e} is (speed of electron =C/100= C/100 where C is the velocity of light)

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