
The mass of radon-222 corresponding to 1curie will be (its half life is 3.8days)
A. $6.46\times {{10}^{-8}}gm$
B. $64.60\times {{10}^{-7}}gm$
C. $6.46\times {{10}^{-10}}gm$
D. $64.60\times {{10}^{-10}}gm$
Answer
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Hint: Firstly we could convert the given half life that is in days into seconds. Then, we could recall a relation between the half life and the number of atoms left after decay. Now we could find the mass of one radon atom from the given atomic mass and then multiply that value with the above found number of atoms to get the answer.
Formula used:
Half life,
${{T}_{\dfrac{1}{2}}}=0.693\dfrac{N}{n}$
Complete answer:
In the question, we are asked to find the mass of radon-222 corresponding to 1curie. We are also given a half-life to be 3.8days.
${{T}_{\dfrac{1}{2}}}=3.8\times 60\times 60\times 24=328320s$
The activity of radon-222 that corresponds to 1 curie is,
$1curie=3.7\times {{10}^{10}}decay/s$
Now, let us recall a relationship between the number of decays, number of atoms left and the half life. This is given by,
${{T}_{\dfrac{1}{2}}}=0.693\dfrac{N}{n}$
Where, N is the number of atoms remaining and n is the number of decays that took place.
$N=\dfrac{{{T}_{\dfrac{1}{2}}}\times n}{0.693}$
Now, we could substitute the given values to get,
$N=\dfrac{328320s\times 3.7\times {{10}^{10}}}{0.693}=17.53\times {{10}^{5}}atoms$
We know that, by definition the atomic mass of an element contains Avogadro number of particles, that is, 222gm of radon will contain Avogadro number of particles,
${{N}_{A}}=6.022\times {{10}^{23}}atoms\to 222gm$
$\Rightarrow 1atom\to \dfrac{222}{6.022\times {{10}^{23}}}=3.686\times {{10}^{-22}}gm$
Now we could multiply this with the number of atoms that remain after decay to get the answer. That is,
$17.53\times {{10}^{15}}atoms\to 3.686\times {{10}^{-22}}\times 17.53\times {{10}^{15}}=64.60\times {{10}^{-7}}gm$
Therefore we found the mass of radon-222 corresponding to 1curie to be$64.60\times {{10}^{-7}}gm$.
Hence option B is correct.
Note:
Let us understand that Radon222 is considered to be the most stable isotope of radon. It is known to be the immediate decay product of Radon226. It is also known to be the leading cause of lung cancer and the reason has to do with its gaseous nature and also its high radioactivity. Another name is radium emanation.
Formula used:
Half life,
${{T}_{\dfrac{1}{2}}}=0.693\dfrac{N}{n}$
Complete answer:
In the question, we are asked to find the mass of radon-222 corresponding to 1curie. We are also given a half-life to be 3.8days.
${{T}_{\dfrac{1}{2}}}=3.8\times 60\times 60\times 24=328320s$
The activity of radon-222 that corresponds to 1 curie is,
$1curie=3.7\times {{10}^{10}}decay/s$
Now, let us recall a relationship between the number of decays, number of atoms left and the half life. This is given by,
${{T}_{\dfrac{1}{2}}}=0.693\dfrac{N}{n}$
Where, N is the number of atoms remaining and n is the number of decays that took place.
$N=\dfrac{{{T}_{\dfrac{1}{2}}}\times n}{0.693}$
Now, we could substitute the given values to get,
$N=\dfrac{328320s\times 3.7\times {{10}^{10}}}{0.693}=17.53\times {{10}^{5}}atoms$
We know that, by definition the atomic mass of an element contains Avogadro number of particles, that is, 222gm of radon will contain Avogadro number of particles,
${{N}_{A}}=6.022\times {{10}^{23}}atoms\to 222gm$
$\Rightarrow 1atom\to \dfrac{222}{6.022\times {{10}^{23}}}=3.686\times {{10}^{-22}}gm$
Now we could multiply this with the number of atoms that remain after decay to get the answer. That is,
$17.53\times {{10}^{15}}atoms\to 3.686\times {{10}^{-22}}\times 17.53\times {{10}^{15}}=64.60\times {{10}^{-7}}gm$
Therefore we found the mass of radon-222 corresponding to 1curie to be$64.60\times {{10}^{-7}}gm$.
Hence option B is correct.
Note:
Let us understand that Radon222 is considered to be the most stable isotope of radon. It is known to be the immediate decay product of Radon226. It is also known to be the leading cause of lung cancer and the reason has to do with its gaseous nature and also its high radioactivity. Another name is radium emanation.
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