A radioactive isotope X has a half life of 3 seconds. At t=0, a given sample of this isotope contains 8000 atom. Calculate (i) its decay constant (ii) average life (iii) the time `t_1`, when 1000 atoms of the isotope X remain in the sample (iv) number of decay/sec in the sample at `t=t_1sec.`

Options

A. 2s
B. 4s
C. 7s
D. 9s

Answer

Let's solve the problem step by step.

### Given Data:
- Half-life of isotope X, \( T_{1/2} = 3 \) seconds
- Initial number of atoms, \( N_0 = 8000 \)

### (i) Calculate the Decay Constant (\( \lambda \)):
The decay constant can be calculated using the formula:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
Substituting the values:
\[ \lambda = \frac{0.693}{3} = 0.231 \, \text{s}^{-1} \]

### (ii) Calculate the Average Life (\( \tau \)):
The average life is given by the formula:
\[ \tau = \frac{1}{\lambda} \]
Substituting the value of \( \lambda \):
\[ \tau = \frac{1}{0.231} \approx 4.33 \, \text{s} \]

### (iii) Calculate the Time \( t_1 \) when 1000 atoms remain:
We know that the number of atoms remaining at time \( t \) can be expressed as:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
We need to find \( t_1 \) when \( N = 1000 \):
\[ 1000 = 8000 \left( \frac{1}{2} \right)^{\frac{t_1}{3}} \]
Dividing both sides by 8000:
\[ \frac{1000}{8000} = \left( \frac{1}{2} \right)^{\frac{t_1}{3}} \]
This simplifies to:
\[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t_1}{3}} \]
Recognizing that \( \frac{1}{8} = \left( \frac{1}{2} \right)^3 \):
\[ \left( \frac{1}{2} \right)^{\frac{t_1}{3}} = \left( \frac{1}{2} \right)^3 \]
Thus, we can equate the exponents:
\[ \frac{t_1}{3} = 3 \]
Solving for \( t_1 \):
\[ t_1 = 3 \times 3 = 9 \, \text{s} \]

### (iv) Calculate the Number of Decays per Second at \( t = t_1 \):
The number of decays per second (activity) can be calculated using:
\[ \text{Activity} = \lambda \cdot N \]
At \( t = t_1 \), \( N = 1000 \):
\[ \text{Activity} = 0.231 \cdot 1000 = 231 \, \text{decays/s} \]

### Summary of Results:
1. Decay constant \( \lambda = 0.231 \, \text{s}^{-1} \)
2. Average life \( \tau \approx 4.33 \, \text{s} \)
3. Time \( t_1 = 9 \, \text{s} \)
4. Number of decays per second at \( t = t_1 \) is \( 231 \, \text{decays/s} \)