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 | 1.007825 u | 12H | 2.014102 u | 13H | 3.016050 u | 24He | 4.002603 u |
---|---|---|---|---|---|---|---|
36Li | 6.015123 u | 37Li | 7.016004 u | 3070Zn | 69.925325 u | 3482Se | 81.916709 u |
64152Gd | 151.919803 u | 82206Pb | 205.974455 u | 83209Bi | 208.980388 u | 84210Po | 209.982876 u |
A. The nucleus 36Li can emit an alpha particle
B. The nucleus 84210Po can emit a proton
C. Deuteron and alpha particles can undergo complete fusion.
D. The nuclei 3070Zn and 3482Se can undergo complete fusion
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 nucleus can emit an alpha particle.
We know that an alpha particle is 24He.
Therefore we can write the first statement as,
36Li→24He+12H
From this we can see that for the 36Lito emit alpha particle, the mass of 36Lishould be greater than the mass of 24He and 12H, i.e.
M36Li≥M24He+M12H
From the question we have the mass of 36Lias,
M36Li=6.015123u
Mass of 24Heas,
M24He=4.002603u
Mass of 12H as,
M12H=2.014102u
Therefore we have,
M24He+M12H=4.002603+2.014102
⇒M24He+M12H=6.016705u
We know that 6.015123>6.016705
Therefore we have M36Li>M24He+M12H.
Thus we can say that the nucleus of 36Li can emit alpha particles.
Hence the correct answer is option A.
Note:
The second statement says that 84210Po nucleus can emit a proton.
This reaction can be written as,
84210Po→11H+13H+82206Pb
From this we can say that for 84210Po to emit a proton its mass should be greater than the mass of 11H,13H and 82206Pb. i.e.
M84210Po>M11H+M13H+M82206Pb
We know that,
M84210Po=209.982876u
M11H=1.007825u
M13H=3.016050u
M82206Pb=205.974455u
Therefore we get,
M11H+M13H+M82206Pb=1.007825+3.016050+205.974455
⇒M11H+M13H+M82206Pb=209.99833u
Since 209.982876<209.9983, we can say that M84210Po<M11H+M13H+M82206Pb.
Thus we can say that 84210Pocannot 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 and 3482Se can undergo complete fusion.
This reaction can be written as,
3070Zn+3482Se→64152Gd
For this statement to be correct, M3070Zn+M3482Se=M64152Gd
We know,
M3070Zn=69.925325u
M3482Se=81.916709u
M64152Gd=151.919803u
Therefore we can calculate,
M3070Zn+M3482Se=69.925325+81.916709
⇒M3070Zn+M3482Se=151.842034u
Since 151.8420234<151.919803, we can say that M3070Zn+M3482Se<M64152Gd.
Thus the fusion of 3070Zn and 3482Se is not possible.
Hence this option is also incorrect.