A hydrogen molecule and helium atom are moving with the same velocity. Then the ratio of their de Broglie wavelength is

Options

A. `1:1`
B. `1:27`
C. `2:1`
D. `2:3`

Answer

To solve the problem of finding the ratio of the de Broglie wavelengths of a hydrogen molecule (H₂) and a helium atom (He) moving with the same velocity, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) 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: Set up the ratio of the de Broglie wavelengths
Since both the hydrogen molecule and the helium atom are moving with the same velocity, we can denote their wavelengths as:
- \(\lambda_{H_2}\) for the hydrogen molecule,
- \(\lambda_{He}\) for the helium atom.

Using the de Broglie wavelength formula, we can express the wavelengths as:
\[ \lambda_{H_2} = \frac{h}{m_{H_2} v} \]
\[ \lambda_{He} = \frac{h}{m_{He} v} \]

### Step 3: Calculate the ratio of the wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \frac{\frac{h}{m_{H_2} v}}{\frac{h}{m_{He} v}} = \frac{m_{He}}{m_{H_2}} \]
Since \(h\) and \(v\) are constants and cancel out.

### Step 4: Substitute the masses
The mass of a hydrogen molecule (H₂) is approximately 2 atomic mass units (amu), and the mass of a helium atom (He) is approximately 4 amu. Thus:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \frac{m_{He}}{m_{H_2}} = \frac{4 \, \text{amu}}{2 \, \text{amu}} = 2 \]

### Step 5: Final ratio
Therefore, the ratio of their de Broglie wavelengths is:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = 2:1 \]

### Conclusion
The final answer is that the ratio of the de Broglie wavelengths of the hydrogen molecule to the helium atom is \(2:1\).

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