The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: (Atomic masses: He = 4 u, O = 16 u)

The root mean square speed (vrms) of a gas is given by the formula:

$$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$

where R is the universal gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas in kg/mol. However, since we are comparing two gases and R is common, we can work with molar masses in the same unit (like g/mol) because the conversion factor cancels out in ratios.

For helium atoms (He) at 300 K, the molar mass MHe = 4 g/mol (since helium is monatomic). For oxygen molecules (O2), the molar mass MO₂ = 32 g/mol (since oxygen is diatomic, 16 u × 2 = 32 u).

We need the temperature T for oxygen such that its vrms equals the vrms of helium at 300 K. So we set the two expressions equal:

$$ \sqrt{\frac{3R \cdot 300}{M_{\text{He}}}} = \sqrt{\frac{3R \cdot T}{M_{\text{O₂}}}} $$

Since the square roots are equal, we can square both sides to eliminate them:

$$ \frac{3R \cdot 300}{M_{\text{He}}} = \frac{3R \cdot T}{M_{\text{O₂}}} $$

The factor 3R appears on both sides and is non-zero, so we can cancel it:

$$ \frac{300}{M_{\text{He}}} = \frac{T}{M_{\text{O₂}}} $$

Now, solve for T by cross-multiplying:

$$ T = 300 \times \frac{M_{\text{O₂}}}{M_{\text{He}}} $$

Substitute the molar masses:

$$ T = 300 \times \frac{32}{4} $$

Simplify the fraction:

$$ \frac{32}{4} = 8 $$

So,

$$ T = 300 \times 8 = 2400 $$

Therefore, the temperature for oxygen molecules is 2400 K.

Comparing with the options:
A. 600 K
B. 2400 K
C. 1200 K
D. 300 K
The value 2400 K corresponds to option B.

Hence, the correct answer is Option B.