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Worked Examples · Example 12.6

Q.Uranium has two isotopes of masses 235235 and 238238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed? If atomic mass of fluorine is 1919 units, estimate the percentage difference in speeds at any temperature.

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The lighter isotope (235UF6^{235}\text{UF}_6) has a larger average speed because, at the same temperature, the mean kinetic energy per molecule is the same for both isotopes. The percentage difference in speeds is about 0.42%.

Figure 12.5
Figure 12.5

Why speed depends on mass in a gas

The Kinetic Theory of Gases tells us that at a given temperature, all gas molecules have the same average translational kinetic energy. This is a direct consequence of the equipartition of energy — each degree of freedom gets 12kT\frac{1}{2}kT of energy, and for a monatomic gas (or the translational motion of any molecule), the average kinetic energy is 32kT\frac{3}{2}kT.

For a molecule of mass mm and root-mean-square speed vrmsv_{\text{rms}}:

12mvrms2=32kT\frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} kT

So:

vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}

The key insight: at fixed temperature, speed is inversely proportional to the square root of molecular mass. Lighter molecules move faster on average.


Step-by-step solution

1. Identify the molecular masses

Uranium hexafluoride is UF6\text{UF}_6. Each molecule contains one uranium atom and six fluorine atoms. Fluorine has atomic mass 1919 units.

  • For the lighter isotope (235U^{235}\text{U}):

M235=235+6×19=235+114=349 uM_{235} = 235 + 6 \times 19 = 235 + 114 = 349 \text{ u}

  • For the heavier isotope (238U^{238}\text{U}):

M238=238+6×19=238+114=352 uM_{238} = 238 + 6 \times 19 = 238 + 114 = 352 \text{ u}

2. Which has larger average speed?

Since vrms∝1/Mv_{\text{rms}} \propto 1/\sqrt{M}, the molecule with the smaller mass has the larger speed. M235<M238M_{235} < M_{238}, so 235UF6^{235}\text{UF}_6 has the larger average speed.

3. Estimate the percentage difference in speeds

The percentage difference is usually defined relative to the speed of the heavier molecule (or sometimes as a relative difference). We'll compute:

v235−v238v238×100%\frac{v_{235} - v_{238}}{v_{238}} \times 100\%

Since v∝1/Mv \propto 1/\sqrt{M}:

v235v238=M238M235=352349\frac{v_{235}}{v_{238}} = \sqrt{\frac{M_{238}}{M_{235}}} = \sqrt{\frac{352}{349}}

Let's compute this carefully:

352349=1+3349≈1.008595\frac{352}{349} = 1 + \frac{3}{349} \approx 1.008595

Taking square root (using the approximation 1+x≈1+x/2\sqrt{1+x} \approx 1 + x/2 for small xx):

1.008595≈1+0.0085952=1.0042975\sqrt{1.008595} \approx 1 + \frac{0.008595}{2} = 1.0042975

So:

v235v238≈1.0043\frac{v_{235}}{v_{238}} \approx 1.0043

The fractional difference is: …

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