Question
Question: Let \(N_{\beta}\) be the no of \(\beta\) -particles emitted by 1 gm of \(^{24}Na\) radioactive nucle...
Let Nβ be the no of β -particles emitted by 1 gm of 24Na radioactive nuclei (half-life= 15 hrs. ) in 7.5 hrs, Nβ is close to (Avogadro number=6.023×1023/gmole ),
A.1.75×1022
B.6.2×1021
C.7.5×1021
D.1.25×1022
Solution
Hint: We will have to calculate the radioactive decay constant first. Then after finding the no. of atoms in 1 gm of 24Na ,we can easily solve this problem.
Formula used:
N=N0.e−λt
λ=t21ln2
Complete step-by-step solution:
Half life is given as t21=15hrs . If λ be the radioactive decay constant, we have,
λ=t21ln2⇒λ=15ln2=0.0462hrs−1
Now, 24 gm 24Na contains 6.023×1023 number of atoms.
Or, 1 gm 24Na contains
N0=246.023×1023=2.51×1022 number of atoms.
The time of decay is given to be t = 7.5 hrs.
So, putting the values of N0 , λ and t , we obtain,
N=N0.e−λt=1.77×1022
This is the number of atoms that has remained after 7.5 hrs of decay. Hence, the number of atoms that had been decayed is,
Nβ=N0−N=(2.51×1022)−(1.77×1022)=7.4×1021
This is the number of β particles that has been emitted.
So, option C is the correct answer.
Additional information:
A radioactive decay of a certain amount of radioactive substance takes infinite time to come to an end. It means, the process never completes and some radioactivity is always present in the substance.
Note: While calculating λ, know that (ln 2) is the natural logarithm and not the usual log (10-base) we use. Their relation is ln2=2.303×log2. Another point to remember to avoid making mistakes is that N in the formula is not the number of atoms that has decayed, but the number that has remained.