Ratio between molecules having different speeds

[DavideGenoa]

Hi, friends!

1. Homework Statement :
I have found an exercise where one should calculate the ratio $N_{v_{\text{e}}}/N_{v_{\text{rms}}}$ between the number of molecules having escape speed $v_{\text{e}}=\sqrt{(2G M_t)/r}$ (where $M$ is the mass of Earth) and those moving at the root mean square speed $v_{\text{rms}}=\sqrt{(3RT)/M}$ (where $M$ is the molar mass) in the nitrogen found at a height of $150 \text{ km}$ and a temperature of ##1000\text{ K}##.

Homework Equations

I know the Maxwell-Boltzmann distribution of molecular speeds

$f(v)=4\pi\bigg(\frac{M}{2\pi RT }\bigg)^{\frac{3}{2}}e^{-\frac{M}{2RT}v^2}$​

which means that, if we call $N$ the total number of molecules and $\Delta n$ is the number of molecules having the speed in the interval $\Delta v$,

$dn=N f(v)dv.$​

The Attempt at a Solution

The values of $v_{\text{f}}\approx1.1\cdot 10^4\text{ m/s}$ and $v_{\text{qm}}\approx 940\text{ m/s}$ are immediately computed, but I have no idea how to find $N_{v_{\text{e}}}/N_{v_{\text{rms}}}$. I do not know if I can use the Maxwell-Boltzmann distribution nor how to use it, since, with such a continuous distribution, the probability that a molecule has a definite velocity, rather than being in an interval, is zero. I am not able to use, if it is useful in this case, the approximation $\Delta n\approx N f(v)\Delta v$, either.

I $\infty$-ly thank you all!

 

[DEvens]

I think you should read the question again and be sure you are stating it correctly. The number moving at $exactly$ $v_{rms}$ is zero. So the ratio you have stated does not make sense.

For example: You could calculate the number with velocity greater than $v_e$ and divide that by the total number. That is, you could get the fraction of the total that is moving fast enough to escape the Earth. That gives you an estimate of the fraction that will escape at this temperature. To get the number with velocity over $v_e$ you integrate $f(v)$ from $v_e$ to infinity.

Or you could compute the ratio of $f(v)$ at $v_e$ to $f(v)$ at $v_{rms}$. That is, you could compute the ratio of the probability function at the two speeds. Not quite sure how useful such a thing is. But it depends on what the problem statement is actually asking for.

 

[DavideGenoa]

The question is as I have written. There might be a printing error, nevertheless.
I had thought that

$\lim_{h\to 0}\frac{n(v_{\text{e}}+h)-n(v_{\text{e}})}{n(v_{\text{rms}+h})-n(v_{\text{rms}})} =\frac{\frac{dn(v_{\text{e}})}{dv}}{\frac{dn(v_{\text{rms}})}{dv}}=\frac{f(v_{\text{e}})}{f(v_{\text{rms}})}=e^{\frac{M}{2RT}(v_{\text{e}}^2-v_{\text{rms}}^2)}$​

But, taking the values given in the appendix of my text into account, I compute

$e^{\frac{M}{2RT}(v_{\text{e}}^2-v_{\text{rms}}^2)}=\exp\big({\frac{2\cdot14.0067\cdot10^{-3}}{2\cdot8.31\cdot1000}\big(\frac{6.670\cdot10^{-11}\cdot5.97\cdot10^{24}}{6.37\cdot 10^6+150000}-\frac{3\cdot8.31\cdot 1000}{2\cdot14.0067\cdot10^{-3}}\big) }\big)\approx 6.67\cdot 10^{88}$​

where $14.0067\text{ g}$ is the molar mass of $\text{ N}_2$, $5.97\cdot10^{24}\text{ kg}$ is the mass of Earth and $6.37\cdot 10^6\text{ m}$ its radius, but the book's result is $1\cdot10^{-87}$...
There might well be a printing, or translation (my book is the Italian translation), error...