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Question: The half-life of radium is \({\text{1620}}\) year and the atomic weight is \({\text{226}}\) kg per k...

The half-life of radium is 1620{\text{1620}} year and the atomic weight is 226{\text{226}} kg per kilo mol. The number of atoms that will decay from its 1{\text{1}} g sample per second will be (Take Avogadro number NA=6⋅023×1023{N_A} = 6 \cdot 023 \times {10^{23}} atoms per kilo mol).
A.3⋅61×10103 \cdot 61 \times {10^{10}}
B.3⋅11×10153 \cdot 11 \times {10^{15}}
C.3⋅6×10123 \cdot 6 \times {10^{12}}
D.3⋅1×10153 \cdot 1 \times {10^{15}}

Explanation

Solution

The number of atoms that will decay per second is the rate of decay. We can find out this using the radioactive law. All radioactive decays are first order reactions. Hence we can apply the equations of first order reaction to a radioactive decay process.

Complete step by step answer: According to radioactive law, the number of atoms that will decay per second is proportional to the number of radioactive molecules present. The number of atoms that will decay per second is also called the rate of radioactive decay. It is given by,
Rate of radioactive decay, dNdt=λN0\dfrac{{dN}}{{dt}} = \lambda {N_0}
Where, λ\lambda is the decay constant and N0{N_0} is the number of atoms initially present.
We need to calculate dNdt\dfrac{{dN}}{{dt}}.
Given, the half-life of radium ,
t12={t_{\dfrac{1}{2}}} = 1620{\text{1620}} year = 1620×365×24×60×60=5⋅1088×10101620 \times 365 \times 24 \times 60 \times 60 = 5 \cdot 1088 \times {10^{10}}seconds.
For a radioactive decay, the decay constant is given by,
λ=0⋅6921t12\lambda = \dfrac{{0 \cdot 6921}}{{{t_{\dfrac{1}{2}}}}}
Substitute the value of t12{t_{\dfrac{1}{2}}}.
λ=0⋅69215⋅1088×1010=1⋅355×10−11s−1\lambda = \dfrac{{0 \cdot 6921}}{{5 \cdot 1088 \times {{10}^{10}}}} = 1 \cdot 355 \times {10^{ - 11}}{s^{ - 1}}
Now we need to find out N0{N_0}. N0{N_0} is the number of atoms in 1{\text{1}} g of the sample. We can find out this by dividing the weight of the sample (1g{\text{1g}}) by atomic weight of radium and then multiplying by Avogadro number.
Atomic weight of radium =226 = {\text{226}} kg/kmol
Then, N0=1226×6⋅023×1023=2⋅67×1021{N_0} = \dfrac{1}{{226}} \times 6 \cdot 023 \times {10^{23}} = 2 \cdot 67 \times {10^{21}} atoms.
Substitute the obtained values in dNdt=λN0\dfrac{{dN}}{{dt}} = \lambda {N_0}.
dNdt=1⋅355×10−11×2⋅67×1021=3⋅61×1010\dfrac{{dN}}{{dt}} = 1 \cdot 355 \times {10^{ - 11}} \times 2 \cdot 67 \times {10^{21}} = 3 \cdot 61 \times {10^{10}}atoms per second.
Hence the number of atoms that will decay from 1{\text{1}} g sample per second will be 3⋅61×10103 \cdot 61 \times {10^{10}}.
Therefore, the correct option is A.

Note:
The half-life is given in years in this question. This should be converted into seconds before doing the problem. We should be aware of the units while doing these types of questions. Otherwise the answer will go wrong.