A hydrogen atom has an electron of mass m and a proton of mass M, Mm. The reduced mass is defined to be: u= mM/(m+M) Given u as a series in m/M: The 1st order correction term is defined by including the linear term by no higher order terms. If m = M/1836, by what percentage does including the 1st order correction change the estimate u=m?

Question

A hydrogen atom has an electron of mass m and a proton of mass M, M>>m. The reduced mass is defined to be: u= mM/(m+M) Given u as a series in m/M: The 1st order correction term is defined by including the linear term by no higher order terms. If m = M/1836, by what percentage does including the 1st order correction change the estimate u=m?

unklerhaukus · Answer

The geometric series for |x|<1   is

$$\frac1{1-x}=\sum_{k=0}^\infty x^k=1+x+x^2+x^3+...$$

unklerhaukus · Answer

$$u= \frac{mM}{m+M}=\frac{m}{(m/M)+1}=\frac{m}{1-(-m/M)}$$

$$m = M/1836$$
is equivalent to $$m/M=1/1836$$

so that   |x| = |-m/M| < 1

unklerhaukus · Answer

$$u=\frac m{1-(-m/M)}\=m\Big[1+(-m/M)+(-m/M)^2+...\Big]\=$$