A free `^238U` nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance betwen the alpha particle and the recoiling nucleus at a time t after the decay as measured by the passenger is

Options

A. `x+vt`
B. `x-vt`
C. `x
D. depends on the direction of the train.

Answer

To solve the problem, we need to analyze the situation both when the train is stationary and when it is moving with a uniform velocity \( v \).

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- When the \( ^{238}U \) nucleus decays, it emits an alpha particle and recoils. In the stationary frame (when the train is not moving), the separation between the alpha particle and the recoiling nucleus becomes \( x \) at time \( t \).

2. **Analyzing the Moving Train**:
- When the train is moving with a uniform velocity \( v \), we need to consider the effects of this motion on the measurements made by a passenger in the train.

3. **Inertial Frame Consideration**:
- The train is moving at a constant velocity, which means it is an inertial frame of reference. In an inertial frame, the laws of physics (including the behavior of the particles after decay) remain the same as in a stationary frame.

4. **Relative Motion**:
- At the moment of decay, both the alpha particle and the recoiling nucleus will have their velocities determined by the decay process. However, since the train is moving, the passenger will measure the distances differently due to the motion of the train.

5. **Distance Calculation**:
- After the decay, the alpha particle moves away from the recoiling nucleus. The passenger in the moving train will observe the distance between the alpha particle and the recoiling nucleus as:
\[ \text{Distance} = x + vt \]
- Here, \( vt \) accounts for the distance the train (and thus the observer) has moved during the time \( t \).

### Final Answer:
The distance between the alpha particle and the recoiling nucleus at a time \( t \) after the decay, as measured by the passenger, is:
\[ x + vt \]