Question

The correct statement is

$_{1}^{1}H$	1.007825 u	$_{1}^{2}H$	2.014102 u	$_{1}^{3}H$	3.016050 u	$_{2}^{4}He$	4.002603 u
$_{3}^{6}Li$	6.015123 u	$_{3}^{7}Li$	7.016004 u	$_{30}^{70}Zn$	69.925325 u	$_{34}^{82}Se$	81.916709 u
$_{64}^{152}Gd$	151.919803 u	$_{82}^{206}Pb$	205.974455 u	$_{83}^{209}Bi$	208.980388 u	$_{84}^{210}Po$	209.982876 u

A. The nucleus $_{3}^{6}Li$ can emit an alpha particle
B. The nucleus $_{84}^{210}Po$ can emit a proton
C. Deuteron and alpha particles can undergo complete fusion.
D. The nuclei $_{30}^{70}Zn$ and $_{34}^{82}Se$ can undergo complete fusion

Answer 1

Hint: We are given some radioactive elements and their molecular masses. We are also given four different statements and asked to find which of them is correct. By completing the radioactive reactions given in the question and applying the condition for them to happen, we can find the correct statement.

Complete answer:
In the question we are given some radioactive elements and their mass.
Four statements are given and we are which one of them is correct.
The first statement is that $_{3}^{6}Li$ nucleus can emit an alpha particle.
We know that an alpha particle is $_{2}^{4}He$.
Therefore we can write the first statement as,
$_{3}^{6}Li\to _{2}^{4}He+_{1}^{2}H$
From this we can see that for the $_{3}^{6}Li$to emit alpha particle, the mass of $_{3}^{6}Li$should be greater than the mass of $_{2}^{4}He$ and $_{1}^{2}H$, i.e.
${{M}_{_{3}^{6}Li}}\ge {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}$
From the question we have the mass of $_{3}^{6}Li$as,
${{M}_{_{3}^{6}Li}}=6.015123u$
Mass of $_{2}^{4}He$as,
${{M}_{_{2}^{4}He}}=4.002603u$
Mass of $_{1}^{2}H$ as,
${{M}_{_{1}^{2}H}}=2.014102u$
Therefore we have,
${{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}=4.002603+2.014102$
$\Rightarrow {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}=6.016705u$
We know that $6.015123 > 6.016705$
Therefore we have ${{M}_{_{3}^{6}Li}} > {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}$.
Thus we can say that the nucleus of $_{3}^{6}Li$ can emit alpha particles.

Hence the correct answer is option A.

Note:
The second statement says that $_{84}^{210}Po$ nucleus can emit a proton.
This reaction can be written as,
$_{84}^{210}Po\to _{1}^{1}H+_{1}^{3}H+_{82}^{206}Pb$
From this we can say that for $_{84}^{210}Po$ to emit a proton its mass should be greater than the mass of $_{1}^{1}H,_{1}^{3}H\text{ and }_{82}^{206}Pb$. i.e.
${{M}_{_{84}^{210}Po}}>{{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}$
We know that,
${{M}_{_{84}^{210}Po}}=209.982876u$
${{M}_{_{1}^{1}H}}=1.007825u$
${{M}_{_{1}^{3}H}}=3.016050u$
${{M}_{_{82}^{206}Pb}}=205.974455u$
Therefore we get,
${{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}=1.007825+3.016050+205.974455$
$\Rightarrow {{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}=209.99833u$
Since $209.982876 < 209.9983$, we can say that ${{M}_{_{84}^{210}Po}}<{{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}$.
Thus we can say that $_{84}^{210}Po$cannot emit a proton.
Hence option B is incorrect.
We know that Deuteron and alpha particles can’t undergo complete fusion. Hence option c is also incorrect.
According to the fourth statement $_{30}^{70}Zn$ and $_{34}^{82}Se$ can undergo complete fusion.
This reaction can be written as,
$_{30}^{70}Zn+_{34}^{82}Se\to _{64}^{152}Gd$
For this statement to be correct, ${{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}={{M}_{_{64}^{152}Gd}}$
We know,
${{M}_{_{30}^{70}Zn}}=69.925325u$
${{M}_{_{34}^{82}Se}}=81.916709u$
${{M}_{_{_{64}^{152}Gd}}}=151.919803u$
Therefore we can calculate,
${{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}=69.925325+81.916709$
$\Rightarrow {{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}=151.842034u$
Since $151.8420234<151.919803$, we can say that ${{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}<{{M}_{_{64}^{152}Gd}}$.
Thus the fusion of $_{30}^{70}Zn$ and $_{34}^{82}Se$ is not possible.
Hence this option is also incorrect.