Because of this perpendicular velocity component, the electron move in a helical path centered at the positive nucleus too.
We know that the electron experiences a drag force that eventually hold its velocity at terminal velocity,
drag force, \(F_D=Av^2\)
where \(A\) is a constant and \(v\), the electron velocity.
If we factor this force into the electron orbit around a positive charge.
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } +Av^{ 2 }\)
\( (m_{ e }-Ar_{ e })r_{ e }v^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } \)
\( r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }(m_{ e }-Ar_{ e })v^{ 2 } } \)
Compare this with,
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ ec } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ ec } } \)
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
Is seem that the electron has lost mass,
\(m_e>m_{ e }-Ar_{ e }\)
If we take the ratio of the two values,
\(\cfrac{r_{ec}}{r_e}=\cfrac{m_{ e }-Ar_{ e }}{m_{ e }}\)
\(\cfrac{r_{ec}}{r_e}=1-\cfrac{A}{m_e}r_{e}\)
where \(r_e\) is the measured atomic radius of hydrogen and \(r_{ec}\) is the calculated value.
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
\(r_{ ec }\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*2*(299792458^2))=1.4090e-15m
Notice that the new terminal velocity, \(2c^2\) has been used.
The measured \(r_{e}\) = 5.3e-11 m
\(\cfrac{r_{ec}}{r_e}\) = 1.4090e-15/5.3e-11=2.658e-5 m
\(A=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}\)
\(A\) = (1-2.658e-5)*9.10938291e-31/5.3e-11=1.719e-20 Nm-2s2
This drag coefficient seems low but for the mass of an electron (9.10938291e-31 kg) it is significant.
