Two ideal diatomic gases have their molar masses in the ratio 1:16 and temperature in the ratio 1 : 4. The average kinetic energy per molecule in the two cases will be in the ratio:

Options

A. `1:16`
B. `1:4`
C. `1:2`
D. `1:64`

Explanation

To solve the problem, we need to find the ratio of the average kinetic energy per molecule of two ideal diatomic gases given their molar masses and temperatures in specific ratios.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy Formula**:
The average kinetic energy (KE) of a gas molecule is given by the formula:
\[ KE = \frac{F}{2} RT \]
where \( F \) is the degree of freedom, \( R \) is the universal gas constant, and \( T \) is the temperature.

2. **Identifying Degrees of Freedom**:
For diatomic gases, the degree of freedom \( F \) is 5 (3 translational + 2 rotational). Since both gases are diatomic, they will have the same degree of freedom.

3. **Setting Up the Equations**:
Let the average kinetic energy of the first gas be \( KE_1 \) and that of the second gas be \( KE_2 \):
\[ KE_1 = \frac{5}{2} RT_1 \]
\[ KE_2 = \frac{5}{2} RT_2 \]

4. **Finding the Ratio of Kinetic Energies**:
To find the ratio of the average kinetic energies, we divide \( KE_1 \) by \( KE_2 \):
\[ \frac{KE_1}{KE_2} = \frac{\frac{5}{2} RT_1}{\frac{5}{2} RT_2} \]
The \( \frac{5}{2} R \) terms cancel out:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]

5. **Using the Given Ratios**:
We are given that the temperatures are in the ratio \( T_1 : T_2 = 1 : 4 \). Therefore:
\[ \frac{T_1}{T_2} = \frac{1}{4} \]

6. **Substituting the Temperature Ratio**:
Now substituting this ratio back into the kinetic energy ratio:
\[ \frac{KE_1}{KE_2} = \frac{1}{4} \]

7. **Conclusion**:
Thus, the average kinetic energy per molecule in the two cases will be in the ratio:
\[ KE_1 : KE_2 = 1 : 4 \]

### Final Answer:
The average kinetic energy per molecule in the two cases will be in the ratio **1 : 4**.