A radioactive mixture initially contains two radioisotopes, P and Q, such that the initial… — JAMB UTME

A radioactive mixture initially contains two radioisotopes, $P$ and $Q$ , such that the initial number of undecayed nuclei of $P$ is $8$ times that of $Q$ . If the half-life of isotope $P$ is $2\text{ hours}$ and the half-life of isotope $Q$ is $6\text{ hours}$ , calculate the time, in hours, after which the number of undecayed nuclei of both isotopes will be equal.

Answer: 9 hours

Answer

The time after which the number of undecayed nuclei of both isotopes will be equal is 9 hours.

By applying the radioactive decay law

N(t) = N_0(1/2)^{t/T_{1/2}}

to both isotopes with initial ratio

N_{P0} = 8N_{Q0}

and equating

N_P(t) = N_Q(t)

, we obtain

8 = 2^{t/3}

, which gives

t = 9\text{ hours}

Step-by-Step Solution

Write the decay equations for isotopes P and Q based on their half-lives

N_P(t) = N_{P0}\left(\frac{1}{2}\right)^{t/2}

and

N_Q(t) = N_{Q0}\left(\frac{1}{2}\right)^{t/6}

Radioactive decay follows the exponential relationship

N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}

Apply the initial condition

N_{P0} = 8 N_{Q0}

and set the two expressions equal

8 N_{Q0} \left(\frac{1}{2}\right)^{t/2} = N_{Q0} \left(\frac{1}{2}\right)^{t/6}

The problem asks for the time

t

when both isotopes have equal remaining undecayed nuclei.

Divide both sides by

N_{Q0} \left(\frac{1}{2}\right)^{t/2}

and simplify the exponents

8 = \frac{(1/2)^{t/6}}{(1/2)^{t/2}} = \left(\frac{1}{2}\right)^{-t/3} = 2^{t/3}

Applying exponent laws simplifies the ratio of powers of one-half into a single base-two exponent.

Solve for time

t

using powers of 2

2^3 = 2^{t/3} \implies \frac{t}{3} = 3 \implies t = 9\text{ hours}

Equating the exponents of identical base 2 gives the exact time.

Key Concept

Radioactive Decay Law and Half-life for Isotope Mixtures

Estimated Time:2m 0s

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