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Q.The waves associated with a moving electron and a moving proton have the same wavelength λ\lambda. It implies that they have the same : (A) momentum (B) angular momentum (C) speed (D) energy

Tripura TbseCBSE Class XII Board 2024MCQ· 1mImportance★★★★★
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De Broglie's relation λ=hp\lambda = \frac{h}{p} shows that equal wavelengths mean equal momenta, regardless of mass. The answer is (A) momentum.

Why wavelength determines momentum

De Broglie's revolutionary insight was that every moving particle has a wave associated with it, with wavelength inversely proportional to its momentum. The relation is beautifully simple:

λ=hp\lambda = \frac{h}{p}

where hh is Planck's constant and pp is the momentum. This is the foundation of wave-particle duality.

Notice what this equation tells us: wavelength depends only on momentum, not on mass, not on kinetic energy, not on speed individually. If two particles—no matter how different their masses—have the same de Broglie wavelength, they must have identical momenta.

Step-by-step analysis

  1. Apply de Broglie's relation to both particles

    For the electron:

λ=hpe\lambda = \frac{h}{p_e}

For the proton:

λ=hpp\lambda = \frac{h}{p_p}

  1. Equate the wavelengths

    Since both wavelengths are equal:

hpe=hpp\frac{h}{p_e} = \frac{h}{p_p}

Canceling hh from both sides:

pe=ppp_e = p_p

So the momenta are identical. This immediately confirms option (A).

  1. Check the other options

    Now let's see why the remaining quantities differ. Remember that a proton is roughly 1836 times heavier than an electron: mp≈1836 mem_p \approx 1836 \, m_e.

    Speed: Since p=mvp = mv, equal momentum means:

meve=mpvpm_e v_e = m_p v_p

ve=mpmevp≈1836 vpv_e = \frac{m_p}{m_e} v_p \approx 1836 \, v_p

The electron moves much faster. Speeds are not equal.

  1. Energy comparison

    For non-relativistic particles, kinetic energy is:

E=p22mE = \frac{p^2}{2m}

With equal momentum pp:

Ee=p22me,Ep=p22mpE_e = \frac{p^2}{2m_e}, \quad E_p = \frac{p^2}{2m_p}

Since mp≫mem_p \gg m_e:

Ee=mpmeEp≈1836 EpE_e = \frac{m_p}{m_e} E_p \approx 1836 \, E_p …

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