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Question: Consider the radioactive decay, X(radioactive) $\longrightarrow$ Y(stable). At time $t=0$, X is pres...

Consider the radioactive decay, X(radioactive) ⟶\longrightarrow Y(stable). At time t=0t=0, X is present in the pure form and at t=2t=2 h, the ratio of amounts of X and Y is 1:3. The duration required for the ratio to become 1:15 is

A

10 h

B

8 h

C

6 h

D

4 h

Answer

4 h

Explanation

Solution

The problem describes a first-order radioactive decay process: X (radioactive) ⟶\longrightarrow Y (stable).

Let N0N_0 be the initial amount of X at t=0t=0. At any time tt, the amount of radioactive substance X remaining is given by the formula for first-order decay:

NX(t)=N0e−λtN_X(t) = N_0 e^{-\lambda t}

where λ\lambda is the decay constant.

The amount of stable product Y formed at time tt is the difference between the initial amount of X and the amount of X remaining:

NY(t)=N0−NX(t)=N0−N0e−λt=N0(1−e−λt)N_Y(t) = N_0 - N_X(t) = N_0 - N_0 e^{-\lambda t} = N_0 (1 - e^{-\lambda t})

The ratio of the amounts of X and Y at time tt is:

NX(t)NY(t)=N0e−λtN0(1−e−λt)=e−λt1−e−λt\frac{N_X(t)}{N_Y(t)} = \frac{N_0 e^{-\lambda t}}{N_0 (1 - e^{-\lambda t})} = \frac{e^{-\lambda t}}{1 - e^{-\lambda t}}

Step 1: Use the given information at t=2t=2 h to find a relationship involving λ\lambda.

At t=2t=2 h, the ratio of amounts of X and Y is 1:3.

NX(2)NY(2)=13\frac{N_X(2)}{N_Y(2)} = \frac{1}{3}

Substitute t=2t=2 into the ratio formula:

e−λ(2)1−e−λ(2)=13\frac{e^{-\lambda (2)}}{1 - e^{-\lambda (2)}} = \frac{1}{3}

3e−2λ=1−e−2λ3e^{-2\lambda} = 1 - e^{-2\lambda}

4e−2λ=14e^{-2\lambda} = 1

e−2λ=14e^{-2\lambda} = \frac{1}{4}

Step 2: Determine the time t′t' when the ratio becomes 1:15.

We need to find the duration t′t' for which NX(t′)NY(t′)=115\frac{N_X(t')}{N_Y(t')} = \frac{1}{15}.

Using the ratio formula:

e−λt′1−e−λt′=115\frac{e^{-\lambda t'}}{1 - e^{-\lambda t'}} = \frac{1}{15}

15e−λt′=1−e−λt′15e^{-\lambda t'} = 1 - e^{-\lambda t'}

16e−λt′=116e^{-\lambda t'} = 1

e−λt′=116e^{-\lambda t'} = \frac{1}{16}

Step 3: Solve for t′t' using the relationship from Step 1.

From Step 1, we have e−2λ=14e^{-2\lambda} = \frac{1}{4}.

This can be rewritten as (e−λ)2=14(e^{-\lambda})^2 = \frac{1}{4}.

Taking the square root of both sides (and knowing e−λe^{-\lambda} must be positive):

e−λ=14=12e^{-\lambda} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Now substitute this into the equation from Step 2:

e−λt′=116e^{-\lambda t'} = \frac{1}{16}

(e−λ)t′=116(e^{-\lambda})^{t'} = \frac{1}{16}

Substitute e−λ=12e^{-\lambda} = \frac{1}{2}:

(12)t′=116\left(\frac{1}{2}\right)^{t'} = \frac{1}{16}

Since 16=2416 = 2^4, we can write 116\frac{1}{16} as (12)4\left(\frac{1}{2}\right)^4.

(12)t′=(12)4\left(\frac{1}{2}\right)^{t'} = \left(\frac{1}{2}\right)^4

Therefore, t′=4t' = 4 h.

The duration required for the ratio to become 1:15 is 4 hours.