Question

{"given":"A particle of mass $M$ at rest decays: $M \\rightarrow m_1 + m_2$. Initial momentum $= 0$.","key_observation":"By conservation of momentum, $\\vec{p}_1 + \\vec{p}_2 = 0$, so $|p_1| = |p_2| = p$. The de Broglie wavelength is $\\lambda = \\dfrac{h}{p}$, which depends only on the magnitude of momentum.","option_analysis":[{"label":"(A)","text":"m1/m2","verdict":"incorrect","explanation":"This ratio would apply if both particles had equal kinetic energies, not equal momenta. Since $|p_1| = |p_2|$, the mass ratio is irrelevant to the wavelength ratio."},{"label":"(B)","text":"m2/m1","verdict":"incorrect","explanation":"This is also a mass-ratio distractor and does not follow from conservation of momentum. Equal momenta give equal wavelengths."},{"label":"(C)","text":"1","verdict":"correct","explanation":"Since $|p_1| = |p_2| = p$, we get $\\lambda_1 = \\dfrac{h}{p_1} = \\dfrac{h}{p_2} = \\lambda_2$, so $\\dfrac{\\lambda_1}{\\lambda_2} = 1$."},{"label":"(D)","text":"√m2 / √m1","verdict":"incorrect","explanation":"This ratio $\\sqrt{m_2}/\\sqrt{m_1}$ would be correct if both particles had equal kinetic energies (since $\\lambda = h/\\sqrt{2mK}$), but here they have equal momenta, not equal kinetic energies."}],"answer":"(C)","formula_steps":[]}