Question

$ 1.0g $ of $ _{79}A{u^{198}} $ (half-life $ = 65hr $ ) decays by $ \beta - $ emission to produce stable $ Hg $ $ {(_{80}}H{g^{198}}) $ . how much $ Hg $ ( in grams) will be present after $ 260hr? $
(A) $ \dfrac{{16}}{{15}}g $
(B) $ \dfrac{{15}}{{16}}g $
(C) $ \dfrac{{21}}{{20}}g $
(D) $ \dfrac{{20}}{{21}}g $

Answer 1

Hint: We know that beta decay is a type of radioactive decay in which a proton is transformed into a neutron or vice versa inside the nucleus of the atom. This means that either a neutron can be converted into a proton or a proton into a neutron. Also, there are two types of beta decay which are beta minus $ (\beta - ) $ and beta plus $ (\beta + ) $ .

Complete Step By Step Answer:
The nuclear reaction for the process will be:
 $ _{79}A{u^{198}}{ \to _{80}}H{g^{198}}{ + _{ - 1}}{\beta ^0} $
As we see that there is an increase in the atomic number of the atom so, this reaction will be termed as beta – minus decay. In this reaction what happens is that a neutron is transformed to yield a proton which causes an increase in the atomic number of the atom. The neutron is neutral but the proton is positive.
Now let us find the amount of $ Hg $ that will be present after $ 260hrs $
So, the half – life $ {t_{\left( {\dfrac{1}{2}} \right)}} = 65hr $ $ = 65hr $
Total time $ (T) = 260hr $
We know that $ T = {t_{\left( {\dfrac{1}{2}} \right)}} \times n $
So,
 $ n = \dfrac{{{t_{\left( {\dfrac{1}{2}} \right)}}}}{T} $
 $ = \dfrac{{260}}{{65}} $
 $ = 4 $
Now the amount which will be left after the radioactive decay will be:
 $ {N_t} = \dfrac{{{N_0}}}{{{2^n}}} $
 $ = \dfrac{1}{{{2^4}}} $
 $ = \dfrac{1}{{16}}g $
So, the amount of $ Au $ decayed will be $ = \dfrac{{15}}{{16}}g $ and therefore the amount of $ Hg $ formed will also be $ = \dfrac{{15}}{{16}}g $ $ $
Therefore, the correct answer is Option B.

Note:
In beta decay reactions in order to maintain the conservation of charge, the nucleus in the process also produces an electron and an antineutrino. Antineutrino is the antimatter counterpart of neutrino. The beta particles emitted after the decay can cause burns and damage to the internal cell or organs if inhaled.