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Question: Let \(N_{\beta}\) be the no of \(\beta\) -particles emitted by 1 gm of \(^{24}Na\) radioactive nucle...

Let NβN_{\beta} be the no of β\beta -particles emitted by 1 gm of 24Na^{24}Na radioactive nuclei (half-life= 15 hrs. ) in 7.5 hrs, NβN_{\beta} is close to (Avogadro number=6.023×1023/gmole6.023×10^{23} /g mole ),

A.1.75×1022A. 1.75\times {{10}^{22}}
B.6.2×1021B. 6.2\times {{10}^{21}}
C.7.5×1021C. 7.5\times {{10}^{21}}
D.1.25×1022D. 1.25\times {{10}^{22}}

Explanation

Solution

Hint: We will have to calculate the radioactive decay constant first. Then after finding the no. of atoms in 1 gm of 24Na^{24} Na ,we can easily solve this problem.

Formula used:
N=N0.eλtN=N_0 .e^{-\lambda t}
λ=ln2t12\lambda =\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}}

Complete step-by-step solution:
Half life is given as t12=15hrs{{t}_{\dfrac{1}{2}}}=15hrs . If λ\lambda be the radioactive decay constant, we have,

λ=ln2t12 λ=ln215=0.0462hrs1 \begin{aligned} & \lambda =\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}} \\\ & \Rightarrow \lambda =\dfrac{\ln 2}{15}=0.0462hr{{s}^{-1}} \\\ \end{aligned}

Now, 24 gm 24Na^{24}Na contains 6.023×10236.023\times {{10}^{23}} number of atoms.

Or, 1 gm 24Na^{24}Na contains

N0=6.023×102324=2.51×1022{{N}_{0}}=\dfrac{6.023\times {{10}^{23}}}{24}=2.51\times {{10}^{22}} number of atoms.
The time of decay is given to be t = 7.5 hrs.

So, putting the values of N0N_0 , λ\lambda and t , we obtain,

N=N0.eλt=1.77×1022N={{N}_{0}}.{{e}^{-\lambda t}}=1.77\times {{10}^{22}}

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=N0N=(2.51×1022)(1.77×1022)=7.4×1021{{N}_{\beta }}={{N}_{0}}-N=(2.51\times {{10}^{22}})-(1.77\times {{10}^{22}})=7.4\times {{10}^{21}}

This is the number of β\beta 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\lambda, know that (ln 2) is the natural logarithm and not the usual log (10-base) we use. Their relation is ln2=2.303×log2\ln 2=2.303\times \log 2. 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.