Predict which of the following statements are true or false.
The de-Broglie wavelength associated with hydrogen atom is large than that of deuterium atom if both are travelling at the same speed.

Answer

To determine whether the statement "The de-Broglie wavelength associated with hydrogen atom is larger than that of deuterium atom if both are traveling at the same speed" is true or false, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Analyze the given conditions
In this case, we are told that both the hydrogen atom and the deuterium atom are traveling at the same speed (\( v \)). Therefore, we can conclude that the velocity (\( v \)) is constant for both atoms.

### Step 3: Compare the masses of hydrogen and deuterium
- The mass of a hydrogen atom (protium) is approximately \( 1 \, u \) (atomic mass unit).
- The mass of a deuterium atom is approximately \( 2 \, u \).

### Step 4: Relate the de-Broglie wavelength to mass
Since the velocity is constant, we can see that the de-Broglie wavelength is inversely proportional to the mass of the atom:
\[ \lambda \propto \frac{1}{m} \]
This means that as the mass increases, the wavelength decreases.

### Step 5: Draw conclusions based on the masses
Since the mass of deuterium is greater than the mass of hydrogen:
- \( m_{\text{deuterium}} > m_{\text{hydrogen}} \)
- Therefore, \( \lambda_{\text{deuterium}} < \lambda_{\text{hydrogen}} \)

### Final Conclusion
The de-Broglie wavelength associated with the hydrogen atom is indeed larger than that of the deuterium atom when both are traveling at the same speed. Thus, the statement is **True**.

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