Question

Radioactive material 'A' has decay constant '8λ' and material 'B' has decay constant 'λ'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be \(\frac 1e\) ?

• A. \(\frac 1λ\)
• B. \(\frac {1}{7λ}\)
• C. \(\frac {1}{8λ}\)
• D. \(\frac {1}{9λ}\)

Answer 1

The Correct Option is B

To solve this problem, we start by using the formula for radioactive decay. The number of undecayed nuclei at any time \( t \) is given by:

N(t) = N_0 e^{-\lambda t}

where \( N_0 \) is the initial number of nuclei and \( \lambda \) is the decay constant.

For material 'A', the decay constant is \( 8\lambda \), so the number of undecayed nuclei at time \( t \) is:

N_A(t) = N_0 e^{-8\lambda t}

For material 'B', the decay constant is \( \lambda \), so the number of undecayed nuclei at time \( t \) is:

N_B(t) = N_0 e^{-\lambda t}

We want to find the time \( t \) at which the ratio of the number of nuclei of material 'B' to material 'A' is \( \frac{1}{e} \).

The ratio is given by:

\(\frac{N_B(t)}{N_A(t)} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-8\lambda t}} = e^{7\lambda t}\)

Set this equal to \( \frac{1}{e} \):

e^{7\lambda t} = e^{-1}

Taking natural logarithms on both sides, we get:

7\lambda t = -1

Solving for \( t \), we find:

t = -\frac{1}{7\lambda}

Since time cannot be negative in this context, we should consider the positive time interval, which gives the correct option as:

t = \frac{1}{7\lambda}

Thus, the correct answer is \(\frac{1}{7\lambda}\), meaning the ratio of nuclei of material 'B' to 'A' becomes \( \frac{1}{e} \) after \( \frac{1}{7\lambda} \) units of time.