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The ratio of de-Broglie wavelength of a proton and an α-particle accelerated through the same potential difference is.
    Correct answer is '2.828'. Can you explain this answer?
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    The correct answer is: 2.828
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    The ratio of de-Broglie wavelength of a proton and anα-particle ...
    Electron can be calculated using the formula:

    λ = h / p

    where λ is the de-Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.

    The momentum of a particle can be calculated using the formula:

    p = mv

    where m is the mass of the particle and v is its velocity.

    The mass of a proton is approximately 1.6726219 x 10^-27 kg, and the mass of an electron is approximately 9.10938356 x 10^-31 kg.

    The ratio of the de-Broglie wavelength of a proton to that of an electron can be calculated as follows:

    λ_proton / λ_electron = (h / p_proton) / (h / p_electron)

    Since h is common to both terms, it cancels out:

    λ_proton / λ_electron = p_electron / p_proton

    Using the formulas for momentum, we have:

    λ_proton / λ_electron = (m_electron * v_electron) / (m_proton * v_proton)

    Plugging in the values for the masses and velocities of the proton and electron, we get:

    λ_proton / λ_electron = [(9.10938356 x 10^-31 kg) * v_electron] / [(1.6726219 x 10^-27 kg) * v_proton]

    Simplifying further, we have:

    λ_proton / λ_electron = 5.446 × 10^-4 * (v_electron / v_proton)

    Therefore, the ratio of the de-Broglie wavelength of a proton to that of an electron is approximately 5.446 × 10^-4 times the ratio of their velocities.
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    The ratio of de-Broglie wavelength of a proton and anα-particle ...

     

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    The correct answer is: 2.828
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