The kinetic energy of `alpha`- particles emiited in the decay of `._(88)Ra^(226)` into `._(86)Rn^(222)` is measured to be `4.78 MeV`. What is the total disintegration energy or the `'Q'`-value of this process ?

Answer

To find the total disintegration energy or the 'Q'-value of the decay process of radium-226 (Ra-226) to radon-222 (Rn-222) with the emission of an alpha particle, we can follow these steps:

### Step 1: Understand the decay process
In the decay of radium-226, an alpha particle (which is a helium nucleus, consisting of 2 protons and 2 neutrons) is emitted. The decay can be represented as:

\[ _{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He} \]

### Step 2: Identify the kinetic energy of the alpha particle
The kinetic energy (K) of the emitted alpha particle is given as 4.78 MeV.

### Step 3: Relate kinetic energy to the Q-value
The Q-value of a nuclear decay process is the total energy released during the decay. It can be related to the kinetic energy of the emitted particles. For alpha decay, the relationship can be expressed as:

\[ Q = K_{\alpha} + K_{Rn} \]

where \( K_{\alpha} \) is the kinetic energy of the alpha particle and \( K_{Rn} \) is the kinetic energy of the daughter nucleus (Rn-222). However, since Rn-222 is much heavier than the alpha particle, we can often neglect its kinetic energy for simplicity. Thus, we can approximate:

\[ Q \approx K_{\alpha} \]

### Step 4: Calculate the Q-value
Since we only have the kinetic energy of the alpha particle, we can directly use this value as an approximation for the Q-value:

\[ Q \approx K_{\alpha} = 4.78 \text{ MeV} \]

### Step 5: Finalize the answer
Thus, the total disintegration energy or the Q-value of the decay process is approximately:

\[ Q \approx 4.78 \text{ MeV} \]

### Summary
The total disintegration energy or Q-value for the decay of radium-226 into radon-222 with the emission of an alpha particle is approximately 4.78 MeV.