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):}|`
Taking kinetic energy ( in `KeV`) of the alpha particle, when the nucleus `._(84)^(210)P_(0)` at rest undergoes alpha decay, is:

Options

A. a.`5319`
B. b.`5422`
C. c.`5707`
D. d.`5818`

Explanation

To solve the problem of finding the kinetic energy of the alpha particle when the nucleus \( _{84}^{210}Po \) undergoes alpha decay, we will follow these steps:

### Step 1: Write the Alpha Decay Reaction
The alpha decay of \( _{84}^{210}Po \) can be represented as:
\[ _{84}^{210}Po \rightarrow _{2}^{4}He + _{82}^{206}Pb \]
This indicates that the polonium nucleus emits an alpha particle (helium nucleus) and transforms into lead.

### Step 2: Determine the Masses
From the given table, we need to find the masses of the involved nuclei:
- Mass of \( _{84}^{210}Po \) = 209.982876 u
- Mass of \( _{2}^{4}He \) = 4.002603 u
- Mass of \( _{82}^{206}Pb \) = 205.974455 u

### Step 3: Calculate the Mass Defect
The mass defect (\( \Delta m \)) can be calculated using the formula:
\[ \Delta m = \text{mass of } _{84}^{210}Po - (\text{mass of } _{2}^{4}He + \text{mass of } _{82}^{206}Pb) \]
Substituting the values:
\[ \Delta m = 209.982876 \, u - (4.002603 \, u + 205.974455 \, u) \]
\[ \Delta m = 209.982876 \, u - 209.977058 \, u = 0.005818 \, u \]

### Step 4: Convert Mass Defect to Energy
The energy equivalent of the mass defect can be calculated using the formula:
\[ E = \Delta m \times 931.5 \, \text{MeV/u} \]
Substituting the mass defect:
\[ E = 0.005818 \, u \times 931.5 \, \text{MeV/u} \approx 5.42 \, \text{MeV} \]

### Step 5: Calculate Kinetic Energy of the Alpha Particle
The kinetic energy of the alpha particle can be found using the formula:
\[ K_{\alpha} = \frac{A_{\alpha}}{A_{parent}} \times E \]
Where:
- \( A_{\alpha} = 4 \) (mass number of the alpha particle)
- \( A_{parent} = 210 \) (mass number of polonium)

Substituting the values:
\[ K_{\alpha} = \frac{4}{210} \times 5.42 \, \text{MeV} \approx 0.1038 \, \text{MeV} \approx 103.8 \, \text{keV} \]

### Final Answer
The kinetic energy of the alpha particle when the nucleus \( _{84}^{210}Po \) undergoes alpha decay is approximately **103.8 keV**.

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