Ratio of the de Broglie wavelenght of molecules of helium and hydrogen at temperatur `27^(@)` C and `327^(@)` C respectively is

Options

A. (a) 2
B. (b) 1
C. (c) 3
D. (d) 4

Answer

To find the ratio of the de Broglie wavelength of helium and hydrogen molecules at temperatures of 27°C and 327°C respectively, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
- The temperature of helium (T₁) is 27°C. To convert to Kelvin:
\[ T₁ = 27 + 273 = 300 \, K \]
- The temperature of hydrogen (T₂) is 327°C. To convert to Kelvin:
\[ T₂ = 327 + 273 = 600 \, K \]

### Step 2: Write the formula for de Broglie wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum. For a gas molecule, the momentum can be expressed in terms of mass (m) and temperature (T):
\[ p = \sqrt{2m k T} \]
where \( k \) is the Boltzmann constant.

### Step 3: Substitute the expression for momentum into the wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2m k T}} \]

### Step 4: Analyze the relationship between wavelength and mass and temperature
From the formula, we can see that the wavelength is inversely proportional to the square root of the product of mass and temperature:
\[ \lambda \propto \frac{1}{\sqrt{mT}} \]

### Step 5: Set up the ratio of the wavelengths for helium and hydrogen
Let \( \lambda_{He} \) be the wavelength of helium and \( \lambda_{H_2} \) be the wavelength of hydrogen. The ratio can be expressed as:
\[ \frac{\lambda_{He}}{\lambda_{H_2}} = \sqrt{\frac{m_{H_2} T_{H_2}}{m_{He} T_{He}}} \]

### Step 6: Substitute the known values
- The molar mass of hydrogen (H₂) is approximately 2 g/mol.
- The molar mass of helium (He) is approximately 4 g/mol.
- Substitute the temperatures and masses into the ratio:
\[ \frac{\lambda_{He}}{\lambda_{H_2}} = \sqrt{\frac{2 \, \text{g/mol} \times 600 \, K}{4 \, \text{g/mol} \times 300 \, K}} \]

### Step 7: Simplify the expression
\[ = \sqrt{\frac{1200}{1200}} = \sqrt{1} = 1 \]

### Conclusion
The ratio of the de Broglie wavelengths of helium and hydrogen at the given temperatures is:
\[ \frac{\lambda_{He}}{\lambda_{H_2}} = 1 \]

### Final Answer
The ratio of the de Broglie wavelength of molecules of helium and hydrogen at temperatures 27°C and 327°C respectively is **1**.

---