The mass of a nucleus `._(Z)^(A)X` is less that the sum of the masses of `(A-Z)` number of neutrons and `Z` number of protons in the nucleus.The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass `M` can break into two light nuclei of masses `m_(1)` and `m_(2)` only if `(m_(1)+m_(2)) lt M`. Also two light nuclei of masses `m_(3)` and `m_(4)` can undergo complete fusion and form a heavy nucleus of mass M'. only if `(m_(3)+m_(4)) gt M'`. The masses of some neutral atoms are given in the table below:
`|{:(._(1)^(1)H ,1.007825u , ._(1)^(2)H,2.014102u,._(1)^(3)H,3.016050u,._(2)^(4)He,4.002603u),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,._(30)^(70)Zn,69.925325u, ._(34)^(82)Se,81.916709u),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}|`
The correct statement is:

Options

A. The nucleus `._(3)^(6)Li` can emit an alpha particle
B. The nucleus `._(84)^(210)P_(0)` can emit a proton
C. Deuteron and alpha particle can undergo complete fusion.
D. The nuclei `._(30)^(70)Zn` and `._(34)^(82)Se` can undergo complete fusion.

Explanation

To solve the problem, we need to analyze the statements provided and determine which one is correct based on the conditions for nuclear reactions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Binding Energy Concept
The binding energy of a nucleus is defined as the energy equivalent to the mass difference between the mass of the nucleus and the sum of the masses of its constituent protons and neutrons. This mass difference arises due to the strong nuclear force that holds the nucleus together.

### Step 2: Analyze the Conditions for Nuclear Reactions
1. For a heavy nucleus of mass \( M \) to break into two lighter nuclei of masses \( m_1 \) and \( m_2 \), the condition is:
\[ m_1 + m_2 < M \]
This means that the total mass of the products must be less than the mass of the original nucleus for the reaction to occur.

2. For two light nuclei of masses \( m_3 \) and \( m_4 \) to fuse and form a heavier nucleus of mass \( M' \), the condition is:
\[ m_3 + m_4 > M' \]
This means that the total mass of the reactants must be greater than the mass of the product for the fusion to be possible.

### Step 3: Evaluate Each Statement
Now we will evaluate each of the given statements based on the above conditions.

#### Statement A: Lithium can emit an alpha particle.
- Lithium-6 (\( ^6Li \)) emitting an alpha particle (\( ^4He \)) would result in:
\[ ^6Li \rightarrow ^4He + ^2H \]
- Check mass difference:
\[ \text{Mass of } ^6Li = 6.015123 \, u \]
\[ \text{Mass of } ^4He = 4.002603 \, u \]
\[ \text{Mass of } ^2H = 2.014102 \, u \]
\[ \Delta m = 6.015123 - (4.002603 + 2.014102) = 6.015123 - 6.016705 = -0.001582 \, u \]
- Since \( \Delta m < 0 \), this reaction is not possible.

#### Statement B: Polonium-210 can emit a proton.
- Polonium-210 (\( ^{210}Po \)) emitting a proton (\( ^1H \)):
\[ ^{210}Po \rightarrow ^{209}Bi + ^1H \]
- Check mass difference:
\[ \text{Mass of } ^{210}Po = 209.982876 \, u \]
\[ \text{Mass of } ^{209}Bi = 208.980388 \, u \]
\[ \Delta m = 209.982876 - (208.980388 + 1.007825) = 209.982876 - 209.988213 = -0.005337 \, u \]
- Since \( \Delta m < 0 \), this reaction is also not possible.

#### Statement C: Deuteron and alpha particle can fuse to form Lithium-6.
- Deuteron (\( ^2H \)) and alpha particle (\( ^4He \)):
\[ ^2H + ^4He \rightarrow ^6Li \]
- Check mass difference:
\[ \Delta m = (2.014102 + 4.002603) - 6.015123 = 6.016705 - 6.015123 = 0.001582 \, u \]
- Since \( \Delta m > 0 \), this reaction is possible.

#### Statement D: Zinc and Selenium can undergo fusion.
- Zinc-70 (\( ^{70}Zn \)) and Selenium-82 (\( ^{82}Se \)):
\[ ^{70}Zn + ^{82}Se \rightarrow \text{Product} \]
- Check mass difference:
\[ \Delta m = 70 + 82 - \text{mass of product} \]
- The product mass must be calculated, and if it results in a negative value, the reaction is not possible.

### Conclusion
From the evaluations, only **Statement C** is correct. The deuteron and alpha particle can indeed fuse to form Lithium-6.