Solveeit Logo

Question

Question: The correct statement is \(_{1}^{1}H\) | 1.007825 u| \(_{1}^{2}H\) | 2.014102 u| \(_{1}^{3}H\) |...

The correct statement is

11H_{1}^{1}H1.007825 u12H_{1}^{2}H2.014102 u13H_{1}^{3}H3.016050 u24He_{2}^{4}He4.002603 u
36Li_{3}^{6}Li6.015123 u37Li_{3}^{7}Li7.016004 u3070Zn_{30}^{70}Zn69.925325 u3482Se_{34}^{82}Se81.916709 u
64152Gd_{64}^{152}Gd151.919803 u82206Pb_{82}^{206}Pb205.974455 u83209Bi_{83}^{209}Bi208.980388 u84210Po_{84}^{210}Po209.982876 u

A. The nucleus 36Li_{3}^{6}Li can emit an alpha particle
B. The nucleus 84210Po_{84}^{210}Po can emit a proton
C. Deuteron and alpha particles can undergo complete fusion.
D. The nuclei 3070Zn_{30}^{70}Zn and 3482Se_{34}^{82}Se can undergo complete fusion

Explanation

Solution

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 36Li_{3}^{6}Li nucleus can emit an alpha particle.
We know that an alpha particle is 24He_{2}^{4}He.
Therefore we can write the first statement as,
36Li24He+12H_{3}^{6}Li\to _{2}^{4}He+_{1}^{2}H
From this we can see that for the 36Li_{3}^{6}Lito emit alpha particle, the mass of 36Li_{3}^{6}Lishould be greater than the mass of 24He_{2}^{4}He and 12H_{1}^{2}H, i.e.
M36LiM24He+M12H{{M}_{_{3}^{6}Li}}\ge {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}
From the question we have the mass of 36Li_{3}^{6}Lias,
M36Li=6.015123u{{M}_{_{3}^{6}Li}}=6.015123u
Mass of 24He_{2}^{4}Heas,
M24He=4.002603u{{M}_{_{2}^{4}He}}=4.002603u
Mass of 12H_{1}^{2}H as,
M12H=2.014102u{{M}_{_{1}^{2}H}}=2.014102u
Therefore we have,
M24He+M12H=4.002603+2.014102{{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}=4.002603+2.014102
M24He+M12H=6.016705u\Rightarrow {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}=6.016705u
We know that 6.015123>6.0167056.015123 > 6.016705
Therefore we have M36Li>M24He+M12H{{M}_{_{3}^{6}Li}} > {{M}_{_{2}^{4}He}}+{{M}_{_{1}^{2}H}}.
Thus we can say that the nucleus of 36Li_{3}^{6}Li can emit alpha particles.

Hence the correct answer is option A.

Note:
The second statement says that 84210Po_{84}^{210}Po nucleus can emit a proton.
This reaction can be written as,
84210Po11H+13H+82206Pb_{84}^{210}Po\to _{1}^{1}H+_{1}^{3}H+_{82}^{206}Pb
From this we can say that for 84210Po_{84}^{210}Po to emit a proton its mass should be greater than the mass of 11H,13H and 82206Pb_{1}^{1}H,_{1}^{3}H\text{ and }_{82}^{206}Pb. i.e.
M84210Po>M11H+M13H+M82206Pb{{M}_{_{84}^{210}Po}}>{{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}
We know that,
M84210Po=209.982876u{{M}_{_{84}^{210}Po}}=209.982876u
M11H=1.007825u{{M}_{_{1}^{1}H}}=1.007825u
M13H=3.016050u{{M}_{_{1}^{3}H}}=3.016050u
M82206Pb=205.974455u{{M}_{_{82}^{206}Pb}}=205.974455u
Therefore we get,
M11H+M13H+M82206Pb=1.007825+3.016050+205.974455{{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}=1.007825+3.016050+205.974455
M11H+M13H+M82206Pb=209.99833u\Rightarrow {{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}=209.99833u
Since 209.982876<209.9983209.982876 < 209.9983, we can say that M84210Po<M11H+M13H+M82206Pb{{M}_{_{84}^{210}Po}}<{{M}_{_{1}^{1}H}}+{{M}_{_{1}^{3}H}}+{{M}_{_{82}^{206}Pb}}.
Thus we can say that 84210Po_{84}^{210}Pocannot 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 3070Zn_{30}^{70}Zn and 3482Se_{34}^{82}Se can undergo complete fusion.
This reaction can be written as,
3070Zn+3482Se64152Gd_{30}^{70}Zn+_{34}^{82}Se\to _{64}^{152}Gd
For this statement to be correct, M3070Zn+M3482Se=M64152Gd{{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}={{M}_{_{64}^{152}Gd}}
We know,
M3070Zn=69.925325u{{M}_{_{30}^{70}Zn}}=69.925325u
M3482Se=81.916709u{{M}_{_{34}^{82}Se}}=81.916709u
M64152Gd=151.919803u{{M}_{_{_{64}^{152}Gd}}}=151.919803u
Therefore we can calculate,
M3070Zn+M3482Se=69.925325+81.916709{{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}=69.925325+81.916709
M3070Zn+M3482Se=151.842034u\Rightarrow {{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}=151.842034u
Since 151.8420234<151.919803151.8420234<151.919803, we can say that M3070Zn+M3482Se<M64152Gd{{M}_{_{30}^{70}Zn}}+{{M}_{_{34}^{82}Se}}<{{M}_{_{64}^{152}Gd}}.
Thus the fusion of 3070Zn_{30}^{70}Zn and 3482Se_{34}^{82}Se is not possible.
Hence this option is also incorrect.