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 mass and temperature ratios.

### Step-by-Step Solution:

1. **Understanding Average Kinetic Energy**:
The average kinetic energy (KE) per molecule of a gas is given by the formula:
\[ KE = \frac{f}{2} k T \]
where:
- \( f \) is the degrees of freedom of the gas,
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature.

2. **Degrees of Freedom for Diatomic Gases**:
For diatomic gases, the degrees of freedom \( f \) is typically 5 (3 translational and 2 rotational). Therefore, we can denote the average kinetic energy for the two gases as:
\[ KE_1 = \frac{5}{2} k T_1 \quad \text{and} \quad KE_2 = \frac{5}{2} k T_2 \]

3. **Finding the Ratio of Average Kinetic Energies**:
The ratio of the average kinetic energies \( KE_1 \) and \( KE_2 \) can be expressed as:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]

4. **Using Given Ratios**:
According to the problem, the temperature ratio is given as:
\[ \frac{T_1}{T_2} = \frac{1}{4} \]

5. **Calculating the Ratio of Average Kinetic Energies**:
Substituting the temperature ratio into the equation for the ratio of average kinetic energies:
\[ \frac{KE_1}{KE_2} = \frac{1}{4} \]

6. **Conclusion**:
Therefore, 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**.