Question

Half-lives of two radioactive elements $A$ and $B$ are $20\, $ minutes and $40\,$ minutes, respectively. Initially, the samples have equal number of nuclei. After $80\,$ minutes, the ratio of decayed numbers of $A$ and $B$ nuclei will be :

• A. $1 : 16$
• B. $4 : 1$
• C. $1 : 4$
• D. $5 : 4$

Answer 1

The Correct Option is D

To solve the problem of finding the ratio of the decayed numbers of nuclei of elements \(A\) and \(B\), we need to use the concept of radioactive decay and half-life.

The number of undecayed nuclei remaining after a time \(t\) is given by the formula:

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

where:

• N_0 is the initial number of nuclei.
• t is the time elapsed.
• T_{1/2} is the half-life of the substance.

Given:

• The half-life of element \(A\) is \(20\) minutes.
• The half-life of element \(B\) is \(40\) minutes.
• Time elapsed \(t = 80\) minutes.
• Initially, \(N_0\) for both \(A\) and \(B\) is equal.

Let's calculate the remaining undecayed nuclei for both elements after \(80\) minutes.

Step 1: Calculate remaining nuclei for element \(A\)

N_A = N_0 \left(\frac{1}{2}\right)^{\frac{80}{20}} = N_0 \left(\frac{1}{2}\right)^{4} = N_0 \times \frac{1}{16}

Step 2: Calculate remaining nuclei for element \(B\)

N_B = N_0 \left(\frac{1}{2}\right)^{\frac{80}{40}} = N_0 \left(\frac{1}{2}\right)^{2} = N_0 \times \frac{1}{4}

Step 3: Calculate the decayed nuclei for both elements

• Decayed nuclei for \(A\): D_A = N_0 - N_A = N_0 - N_0 \times \frac{1}{16} = N_0 \times \frac{15}{16}
• Decayed nuclei for \(B\): D_B = N_0 - N_B = N_0 - N_0 \times \frac{1}{4} = N_0 \times \frac{3}{4}

Step 4: Calculate the ratio of decayed nuclei

The ratio of the decayed nuclei of \(A\) to \(B\) is:

\text{Ratio} = \frac{D_A}{D_B} = \frac{N_0 \times \frac{15}{16}}{N_0 \times \frac{3}{4}} = \frac{15}{16} \times \frac{4}{3} = \frac{5}{4}

Therefore, the ratio of decayed numbers of \(A\) and \(B\) nuclei is 5 : 4, which matches the correct answer.