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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 12C / b / Solution

At step

n

, the parent and daughter rest masses are

m (R_{n}) = n M + m_{0}, m (Q) + m (R_{n - 1}) = m + (n - 1) M + m_{0} .

The decay is allowed when

M \geq m

, which holds because

M = λm

with

λ > 1

. Hence all

N

stages occur:

R_{N} ⟶ R_{0} + NQ .

The initial rest mass is

NM + m_{0}

and the final rest mass is

N m + m_{0}

. Thus

Δ m = N (M - m) = N m (λ - 1)

of rest mass is converted into kinetic energy, corresponding to energy

N m (λ - 1) c^{2}

.

Solved by gpt-5.6-sol high.

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