Question

The half life of a radioactive substance is 10 minutes. If $n_1$ and $n_2$ are the number of atoms decayed in 20 and 30 minutes respectively, then $n_1 : n_2 = $

• A. 7:8
• B. 1:2
• C. 6:7
• D. 3:4

Hint:

Atoms remaining = $N_0 (1/2)^{t/T}$. Atoms decayed = $N_0$ - Atoms remaining.

Answer 1

The Correct Option is C

The problem revolves around the half-life formula of a radioactive substance. The half-life, denoted as \( T_{1/2} \), is the time taken for half of the radioactive atoms to decay. Here, \( T_{1/2} = 10 \) minutes.

Given \( n_1 \) and \( n_2 \) are the number of atoms decayed in 20 and 30 minutes respectively. We need to find the ratio \( n_1 : n_2 \).

The formula for the remaining atoms after time \( t \) is:

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

Where \( N_0 \) is the initial number of atoms.

Step 1: Calculate remaining atoms at \( t = 20 \) minutes:

\[ N(20) = N_0 \left(\frac{1}{2}\right)^{\frac{20}{10}} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \]

Thus, atoms decayed \( n_1 = N_0 - N(20) = N_0 - \frac{N_0}{4} = \frac{3N_0}{4} \).

Step 2: Calculate remaining atoms at \( t = 30 \) minutes:

\[ N(30) = N_0 \left(\frac{1}{2}\right)^{\frac{30}{10}} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8} \]

Thus, atoms decayed \( n_2 = N_0 - N(30) = N_0 - \frac{N_0}{8} = \frac{7N_0}{8} \).

Step 3: Calculate ratio \( n_1 : n_2 \):

\[ n_1 : n_2 = \left(\frac{3N_0}{4} \right) : \left(\frac{7N_0}{8}\right) \]

To simplify:

\[ n_1 : n_2 = \frac{3}{4} \div \frac{7}{8} = \frac{3}{4} \times \frac{8}{7} = \frac{24}{28} = \frac{6}{7} \]

Thus, the ratio \( n_1 : n_2 \) is 6:7.