A radioactive nuclei has a half life of 693s. The activity of one mole of that nuclei sample is:(Avogadro's number=6.023 x1023 and In(2)=0.693)
2×1010 Bq
3.7×1010 Bq
6.023×1020 Bq
0.5×10-10 Bq
1×1020 Bq
Given:
Step 1: Calculate the Decay Constant (\( \lambda \))
The decay constant is given by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{693} = 0.001 \, \text{s}^{-1} \]
Step 2: Determine the Number of Nuclei (\( N \))
For 1 mole of nuclei:
\[ N = N_A = 6.023 \times 10^{23} \]
Step 3: Compute Activity (\( A \))
Activity is defined as:
\[ A = \lambda N = 0.001 \times 6.023 \times 10^{23} = 6.023 \times 10^{20} \, \text{Bq} \]
Conclusion:
The activity of one mole of the radioactive nuclei is \( 6.023 \times 10^{20} \, \text{Bq} \).
Answer: \(\boxed{C}\)
1. Define variables and given information:
2. Calculate the decay constant (λ):
The decay constant is related to the half-life by the following equation:
\[t_{1/2} = \frac{\ln(2)}{\lambda}\]
\[\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{693 \, s} = 10^{-3} \, s^{-1}\]
3. Calculate the number of nuclei (N) in one mole:
One mole of any substance contains Avogadro's number of particles (in this case, nuclei):
\[N = N_A = 6.023 \times 10^{23}\]
4. Calculate the activity (A):
Activity is defined as the rate of decay and is given by:
\[A = \lambda N\]
\[A = (10^{-3} \, s^{-1})(6.023 \times 10^{23}) = 6.023 \times 10^{20} \, Bq\]
If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
Read More: Nuclei