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In a radioactive decay chain, nucleus X decays to 3 nuclei of Y plus particles and energy (decay constant 2*lambda). Each Y nucleus decays to 2 nuclei of Z plus particles and energy (decay constant lambda). Initially, only X is present with mass M (so initial number of X nuclei N0 = M/m_X). The number of Y nuclei is maximum at t = ln(10)/lambda. Find the mass of Y at this maximum instant. (Let m_Y be the mass of one Y nucleus.)

1. 0.04 m
2. 0.06 m
3. 0.08 m
4. 0.02 m

Correct answer: 0.06 m

Solution

At maximum, dN_Y/dt = 0: 6*lambda*N_X = lambda*N_Y, so N_Y = 6*N_X. At t=ln(10)/lambda, N_X = N0*e^(-2*ln10) = N0/100. N_Y = 6*N0/100 = 0.06*N0. Mass of Y = 0.06 * (m_Y/m_X)*M = 0.06*M if m_Y = m_X (or simply 0.06*m where m represents initial mass M and m_Y ~ m_X). Hence mass = 0.06 m.

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