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De Broglie wavelength of a particle compared with a photon's wavelength

This type of question, asked 8 times: May 2021 S1, 3 May 2023 S1, 16 May 2023 S2, 26 Apr 2025 S1, 26 Apr 2025 S2, 11 Apr 2026 S1, 11 Apr 2026 S2, 13 May 2026 S2. All 8 questions 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

    3 May 2023 S1

    11
    The ratio of the wavelength of a photon of energy 'E' to that of the electron of same energy is (m=m= mass of an electron, c=c= speed of light, h=h= Planck's constant)

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  3. Question 3

    16 May 2023 S2

    10
    Kinetic energy of a proton is equal to energy EE of a photon. Let λ1\lambda_1 be the de Broglie wavelength of proton and λ2\lambda_2 be the wavelength of photon. If λ1λ2∝En\frac{\lambda_1}{\lambda_2} \propto E^n then the value of nn is

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  4. Question 4

    26 Apr 2025 S1

    25
    The wavelength ' λ\lambda ' of a photon and the deBroglie wavelength of an electron have same value. the ratio of kinetic energy of the electron to the energy of a photon is (m=m= mass of electron, c=c= velocity of light, h=h = Planck's constant)

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  5. Question 5

    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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  6. Question 6

    11 Apr 2026 S1

    145
    A photon and an electron have equal energy E. The ratio of λ\lambda(electron) to λ\lambda(photon) is proportional to

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  7. Question 7

    11 Apr 2026 S2

    145
    According to de-Broglie hypothesis, the ratio of the wave-length of a photon and that of an electron having same energy 'E' is (m = mass of electron, c = velocity of light)

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  8. Question 8

    13 May 2026 S2

    144
    The energy of a photon is equal to the kinetic energy of a proton. If λ1\lambda_1 is the de-Broglie wavelength of a proton, λ2\lambda_2 is the wavelength associated with the photon and if the energy of the photon is EE then (λ1/λ2)(\lambda_1/\lambda_2) is proportional to

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