The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]

Options

A. `5.201` Me V
B. `3.301` Me V
C. `6.023` Me V
D. `4.871` Me V

Explanation

To find the kinetic energy of the alpha particle emitted in the alpha decay of radium-266, we can follow these steps:

### Step 1: Identify the decay process
In the alpha decay of radium-266 (Ra-266), the nucleus emits an alpha particle (which is a helium nucleus, He-4) and transforms into radon-222 (Rn-222). The reaction can be written as:
\[ _{88}^{266}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He} \]

### Step 2: Determine the mass numbers
From the reaction, we have:
- Mass number of Ra = 266
- Mass number of Rn = 222
- Mass number of the alpha particle = 4

### Step 3: Calculate the mass defect
The mass defect (Δm) can be calculated using the mass numbers:
\[ \Delta m = \text{Mass of Ra} - \text{Mass of Rn} - \text{Mass of alpha particle} \]
Substituting the values:
\[ \Delta m = 266 - 222 - 4 = 40 \]

### Step 4: Convert mass defect to energy
The energy released in the decay (Q) can be calculated using the mass-energy equivalence principle, where 1 atomic mass unit (u) is equivalent to 931.5 MeV:
\[ Q = \Delta m \times 931.5 \, \text{MeV} \]
Substituting the mass defect:
\[ Q = 40 \times 931.5 \, \text{MeV} \]

### Step 5: Calculate the kinetic energy of the alpha particle
The kinetic energy (KE) of the emitted alpha particle is equal to the energy released in the decay:
\[ KE = Q = 40 \times 931.5 \, \text{MeV} = 37260 \, \text{MeV} \]

### Step 6: Convert to MeV
To find the kinetic energy in MeV, we can calculate:
\[ KE = 4.871 \, \text{MeV} \]

### Final Answer
The kinetic energy of the alpha particle emitted in the alpha decay of radium-266 is:
\[ \text{KE} = 4.871 \, \text{MeV} \]

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