Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures `27^circ` and `127^circC`, respectively.

Options

A. `(sqrt(5)/(3))`
B. `(sqrt(8)/(3))`
C. `sqrt((3)/(5))`
D. `sqrt((3)/(8))`

Answer

To find the ratio of the de Broglie wavelengths of hydrogen and helium molecules at given temperatures, we can follow these steps:

### Step 1: Understand the formula for de Broglie wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2m k T}} \]
where:
- \( h \) = Planck's constant
- \( m \) = mass of the molecule
- \( k \) = Boltzmann's constant
- \( T \) = absolute temperature in Kelvin

### Step 2: Convert temperatures from Celsius to Kelvin
- For hydrogen at \( 27^\circ C \):
\[ T_{H_2} = 27 + 273 = 300 \, K \]
- For helium at \( 127^\circ C \):
\[ T_{He} = 127 + 273 = 400 \, K \]

### Step 3: Identify the masses of the molecules
- The mass of a hydrogen molecule (\( H_2 \)) is approximately \( 2 \, u \) (atomic mass units).
- The mass of a helium atom (\( He \)) is approximately \( 4 \, u \).

### Step 4: Write the expressions for the de Broglie wavelengths
- For hydrogen:
\[ \lambda_{H_2} = \frac{h}{\sqrt{2 m_{H_2} k T_{H_2}}} \]
- For helium:
\[ \lambda_{He} = \frac{h}{\sqrt{2 m_{He} k T_{He}}} \]

### Step 5: Find the ratio of the wavelengths
To find the ratio \( \frac{\lambda_{H_2}}{\lambda_{He}} \):
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \frac{\frac{h}{\sqrt{2 m_{H_2} k T_{H_2}}}}{\frac{h}{\sqrt{2 m_{He} k T_{He}}}} = \frac{\sqrt{2 m_{He} k T_{He}}}{\sqrt{2 m_{H_2} k T_{H_2}}} \]
This simplifies to:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_{H_2} T_{H_2}}} \]

### Step 6: Substitute the values
Substituting the known values:
- \( m_{H_2} = 2 \, u \)
- \( m_{He} = 4 \, u \)
- \( T_{H_2} = 300 \, K \)
- \( T_{He} = 400 \, K \)

We get:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{4 \times 400}{2 \times 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{8}{3}} \]

### Final Answer
Thus, the ratio of the de Broglie wavelengths of hydrogen to helium is:
\[ \frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{8}{3}} \]