The atomic masses of He and Ne are 4 and 20 amu respectively . The value of the de Broglie wavelength of He gas at`-73.^(@)C` is ''M'' times that of the de Broglie wavelength of Ne at `727.^(@)C.` M is

Answer

To solve the problem, we need to find the value of \( M \) which represents the ratio of the de Broglie wavelengths of helium (He) gas at \(-73^\circ C\) and neon (Ne) gas at \(727^\circ C\).

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- For helium at \(-73^\circ C\):
\[ T_{He} = -73 + 273 = 200 \, K \]
- For neon at \(727^\circ C\):
\[ T_{Ne} = 727 + 273 = 1000 \, K \]

2. **Write the Formula for de Broglie Wavelength:**
The de Broglie wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{h}{\sqrt{3m k T}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the gas particle,
- \( k \) is Boltzmann's constant,
- \( T \) is the temperature in Kelvin.

3. **Calculate the de Broglie Wavelengths for He and Ne:**
- For helium:
\[ \lambda_{He} = \frac{h}{\sqrt{3 \cdot m_{He} \cdot k \cdot T_{He}}} \]
- For neon:
\[ \lambda_{Ne} = \frac{h}{\sqrt{3 \cdot m_{Ne} \cdot k \cdot T_{Ne}}} \]

4. **Set Up the Ratio of Wavelengths:**
To find the ratio of the wavelengths:
\[ \frac{\lambda_{He}}{\lambda_{Ne}} = \frac{\frac{h}{\sqrt{3 \cdot m_{He} \cdot k \cdot T_{He}}}}{\frac{h}{\sqrt{3 \cdot m_{Ne} \cdot k \cdot T_{Ne}}}} \]
Simplifying this gives:
\[ \frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{m_{Ne} \cdot T_{Ne}}{m_{He} \cdot T_{He}}} \]

5. **Substitute the Known Values:**
- Mass of helium, \( m_{He} = 4 \, \text{amu} \)
- Mass of neon, \( m_{Ne} = 20 \, \text{amu} \)
- Substitute the temperatures:
\[ \frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{20 \cdot 1000}{4 \cdot 200}} \]

6. **Calculate the Ratio:**
\[ = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5 \]

7. **Conclusion:**
The value of \( M \) is:
\[ M = 5 \]