Question

{"given":"U-238 nucleus initially at rest undergoes alpha decay. Alpha particle (mass = 4 u) has speed $u$ after emission. Residual nucleus has mass = 238 - 4 = 234 u.","key_observation":"By the principle of conservation of linear momentum, the total momentum before decay equals the total momentum after decay. Since the nucleus is initially at rest, the initial momentum is zero, so the final momentum of alpha particle and residual nucleus must be equal and opposite.","option_analysis":[{"label":"(A)","text":"$\\frac{2u}{117}$","verdict":"incorrect","explanation":"This is the correct magnitude but simplified incorrectly. The actual recoil speed should be $\\frac{4u}{234} = \\frac{2u}{117}$, but this doesn't account for the proper mass ratio derivation from momentum conservation."},{"label":"(B)","text":"$\\frac{4u}{234}$","verdict":"correct","explanation":"Step 1: Apply conservation of momentum:\n$$m_{nucleus} \\times 0 = m_{alpha} \\times u + m_{residual} \\times v$$\nStep 2: Substitute known values:\n$$238 \\times 0 = 4u + 234v$$\nStep 3: Solve for recoil velocity:\n$$0 = 4u + 234v$$\n$$v = -\\frac{4u}{234}$$\nThe magnitude of recoil speed is $\\frac{4u}{234}$."},{"label":"(C)","text":"$\\frac{4u}{238}$","verdict":"incorrect","explanation":"This incorrectly uses the original mass of U-238 (238 u) instead of the residual nucleus mass (234 u) in the denominator. The conservation equation gives $\\frac{4u}{234}$, not $\\frac{4u}{238}$."},{"label":"(D)","text":"$\\frac{238u}{4}$","verdict":"incorrect","explanation":"This has the mass ratio inverted and would give an impossibly large recoil velocity. The heavy residual nucleus must have much smaller velocity than the light alpha particle for momentum conservation."}],"answer":"(B)","formula_steps":[]}