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Q.A particle of mass MM at rest splits up into two particles of masses m1m_1 and m2m_2 having non-zero velocities. Calculate the ratio of the de Broglie wavelengths associated with the two particles.

Delhi CbseCBSE Class XII Board 2026Subjective· 2mImportance★★★★★
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When a particle at rest splits into two fragments, momentum conservation forces the fragments to carry equal and opposite momenta. Since de Broglie wavelength is inversely proportional to momentum, equal momentum magnitudes mean equal wavelengths: λ1/λ2=1:1\lambda_1/\lambda_2 = 1:1, independent of the masses.

The key insight here is that momentum conservation links the two fragments in a simple way. When a stationary object breaks apart, the total momentum before (zero) must equal the total momentum after. This constraint, combined with the de Broglie relation, gives us the answer without needing to know the actual velocities.

Why momentum conservation is central

A particle at rest has zero momentum. After the split, the two fragments move in opposite directions. For the vector sum of their momenta to remain zero, we must have:

p⃗1+p⃗2=0\vec{p}_1 + \vec{p}_2 = 0

This means p⃗1=−p⃗2\vec{p}_1 = -\vec{p}_2, so the magnitudes are equal:

p1=p2=p(say)p_1 = p_2 = p \quad \text{(say)}

The fragments carry equal momentum magnitudes regardless of their masses. A lighter fragment moves faster, a heavier one slower, but the product mvmv is the same for both.

The de Broglie wavelength

De Broglie's hypothesis assigns a wavelength to any particle with momentum pp:

λ=hp\lambda = \frac{h}{p}

where hh is Planck's constant. The wavelength is inversely proportional to momentum: higher momentum means shorter wavelength.

Finding the ratio

  1. Write the wavelengths for each fragment:

    For particle 1: λ1=hp1\lambda_1 = \dfrac{h}{p_1}

    For particle 2: λ2=hp2\lambda_2 = \dfrac{h}{p_2}

  2. Form the ratio:

λ1λ2=h/p1h/p2=p2p1\frac{\lambda_1}{\lambda_2} = \frac{h/p_1}{h/p_2} = \frac{p_2}{p_1}

  1. Apply momentum conservation: …

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