The ratio of de-Broglie wavelength of molecules of hydrogen and helium gas moving at rms speed in two gas jars kept separately at temperature `37^(@)` C and `127^(@)` C respectively is :

Options

A. `(2)/sqrt(3)`
B. `sqrt(2/(3))`
C. `sqrt(3)/(2)`
D. `sqrt(8/(3)`

Answer

To find the ratio of the de-Broglie wavelength of hydrogen and helium gas molecules moving at their respective root mean square (RMS) speeds at given temperatures, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle. For a gas molecule, momentum can be expressed as:
\[ p = mv \]
where \(m\) is the mass of the molecule and \(v\) is its velocity.

### Step 2: Relate velocity to temperature
For gas molecules, the RMS speed (\(v_{rms}\)) is given by:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature in Kelvin, and \(m\) is the mass of the gas molecule.

### Step 3: Substitute \(v_{rms}\) into the de-Broglie wavelength formula
Substituting \(v_{rms}\) into the momentum expression, we get:
\[ p = m v_{rms} = m \sqrt{\frac{3kT}{m}} = \sqrt{3mkT} \]
Thus, the de-Broglie wavelength becomes:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

### Step 4: Write the ratio of the de-Broglie wavelengths for hydrogen and helium
Let \(m_H\) and \(m_{He}\) be the masses of hydrogen and helium molecules respectively, and \(T_H\) and \(T_{He}\) be their respective temperatures in Kelvin. The ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_H}{\lambda_{He}} = \frac{\frac{h}{\sqrt{3m_H k T_H}}}{\frac{h}{\sqrt{3m_{He} k T_{He}}}} = \frac{\sqrt{3m_{He} k T_{He}}}{\sqrt{3m_H k T_H}} = \frac{\sqrt{m_{He} T_{He}}}{\sqrt{m_H T_H}} \]

### Step 5: Substitute known values
- Mass of hydrogen (\(m_H\)) = 2 amu (atomic mass units)
- Mass of helium (\(m_{He}\)) = 4 amu
- Temperature of hydrogen (\(T_H\)) = 37°C = 310 K
- Temperature of helium (\(T_{He}\)) = 127°C = 400 K

Now substituting these values:
\[ \frac{\lambda_H}{\lambda_{He}} = \frac{\sqrt{4 \times 400}}{\sqrt{2 \times 310}} = \frac{\sqrt{1600}}{\sqrt{620}} = \frac{40}{\sqrt{620}} \]

### Step 6: Simplify the ratio
Calculating \(\sqrt{620}\):
\[ \sqrt{620} \approx 24.9 \]
Thus,
\[ \frac{40}{\sqrt{620}} \approx \frac{40}{24.9} \approx 1.61 \]

### Final Result
The ratio of the de-Broglie wavelengths of hydrogen and helium is approximately:
\[ \frac{\lambda_H}{\lambda_{He}} \approx 1.61 \]