How far the radius and orbital velocity of orbital electron change in helium atom when the electrons are in different orbits? When they are in the same orbit 2e^2/Krnn^2 - e^2/4Krnn^2 = (7/4)e^2/K=(me)nn vnn^2 rnn nh/2π = (me)nn vnn rnn By solving these two relations we get rnn =(4/7)n^2 ao(1- vnn^2/c^2)^1/2 and vnn = (7/4)e^2/2nhεo When one of the electrons is in orbit n1 and the other electron is in orbit n2 Consider the electron in orbit n1 with radius (rn1)n1-n2 its mass, velocity (mn1)n1-n2 and (vn1)n1-n2 2e^2/K(rn1)^2 - e^2/K(rn1 + rn2)^2 =2e^2/K-e^2/K[1/(1+rn2/rn1)]^2 = mn1vn12rn1[e^2/K][2 - [1/(1+n2^2/n1^2]^2 = [e^2/K](n1^4 +2n2^4 +4n1^2n2^2)/(n1^2 + n2^2)^2 = mn1 vn1^2rn1 and n1h/2π = mn1vn1rn1 By solving these two relations, vn1 = [e^2/2n1hεo](n1^4 +2n2^4 +4n1^2n2^2)/(n1^2 + n2^2)^2 rn1 = n1^2 ao [(n1^2 + n2^2)2/(n1^4 +2n2^4 +4n1^2n2^2)] (1- vn1^2/c^2)^1/2. Consider the electron in orbit n2 with radius (rn2)n1-n2 its mass, velocity (mn2)n1-n2 and (vn2)n1-n2 2e^2/K(rn2)^2 - e^2/K(rn1 + rn2)^2 = mn2 vn2^2/rn2 =2e^2/K- e^2/K[1/(1+rn1/rn2)]^2 = mn2vn2^2rn2 [e^2/K][2 -[1/(1+n1^2/n2^2]^2 = [e^2/K](2n1^4 +n2^4 +4n1^2n2^2)/(n1^2 + n2^2)^2 = mn2vn2^2rn2, n2 h/2π = mn2 vn2 rn2. By solving these two relations, vn2 = [e^2/2n1hεo](2n1^4 +n2^4 +4n1^2n2^2)/(n1^2 + n2^2)^2, and rn2 = n2^2 ao [(n1^2 + n2^2)^2/(2n1^4 +n2^4 +4n1^2n2^2)] (1- vn2^2/c^2)^1/2 knowing the orbital velocity , the binding energy of the orbital electron in n1 and n2 can be estimated from (1/2)mo vn1^2 , (1/2)mo vn2^2 respectively. Total binding energy BEn1-n2 = (1/2)mo[(e^2/2hεo)^2/(n1^2 + n2^2)^4][(n1^4 +2n2^4 +4n1^2n2^2)^2/n1^2+(2n1^4 +n2^4 +4n1^2n2^2)^2/n2^2] = [e^2/2Kao][1/(n1^2 + n2^2)4][(n1^4 +2n2^4 +4n1^2n2^2)^2/n1^2+(2n1^4 +n2^4 +4n1^2n2^2)^2/n2^2] Using this general expression, one can find out the binding energy of helium atom in any of its state. n1= n2= 1 ;[e^2/2Kao] (1/16)[ 98] =(49/8)[e^2/2Kao] = 83.269 eV n1 = 2; n2 = 3 ; [e^2/2Kao] [1/(13)^4] [(322/2)^2 +(257/3)^2] = 15.8316 eV n1= n2 = 2; [e^2/2Kao] [1/(8)^4] [(322/2)^2 +(257/3)^2] = 20.8173 eV BE12 - BE23 = 20.8173 - 15.8316 = 4.9857 eV and λ23→12 = 248.72 nm It represents the first line of Balmer series of helium atom . How the radius of the 1s orbit in helium atom changes in accordance with number of electrons in the orbit. When a single electron is present in the nth orbit of helium atom 2e^2/K =mvn^2rn , n^2h^2/4π^2 = m^2vn^2rn^2 and its radius r1 is ao/2. [NPE = 2e^2/Kr = (1/2) m v^2 + dmc^2; KE = (1/2) mv^2 = e^2/Kr or BE= dmc^2 = e^2/Kr = 4e^2 /2Kao]. rn = n^2 r1 is true only for the electronic structures of the orbits are alike. When two electrons are present in the nth orbit, the radius becomes 2e^2/Krnn^2 - e^2/4Krnn^2 = (7/4)e^2/Krnn^2 = mvnn^2/rnn and nh/2π = mvnn rnn , by solving we get rnn = n^2(4/7)ao. . Increase of radius due to the additional electron dr = 0.5714 ao - 0.5 a0 = 0.0714 ao = ao/14. [ NPE = 4e^2/Kr - e^2/2Kr= (7/2) e^2/Kr = 2(1/2) mv^2 + 2 (dmc^2) ; KE =mv^2 = 2 e^2/Kr -e^2/4Kr = (7/4) e^2/Kr and 2(dm)c^2 = (7/4) e^2/Kr = (49/8)e^2/2Kao] . When three electrons are present in the 1s orbit, the radius would be 2e^2/Kr^2 - [2e^2//3Kr^2]cos3o = e^2/Kr^2 {2 - 1/[(3)^1/2]} = 1.42264 e^2/Kr^2; 1.42264 e^2/K = mv^2 r , h^2/4π^2 = m^2v^2r^2 and r = o.7029 ao. Change of radius dr due to the addition of two electrons = o.7029 - o.5 = o.2029ao = ao/5. Change of radius dr due to the addition of third electron = o.7029 - 0.5714 = o.1315 ao = ao/7.6 ≃ 2(ao/14) Why helium negative ion is not possible? In the previous article we explained the non-existence of negative hydrogen ion. This is true for all kinds of atoms which cannot hold one or more additional electrons over its atomic number. Same explanation holds good for any atomic system with one electron excess over its atomic number. For example, helium has two 1s electrons. The innermost orbit of the helium atom is completely filled with a maximum of two electrons. They are held up in the orbit by its nucleus with 2 units of positive charge. The third electron approaching the helium nucleus is not allowed to revolve in any orbits surrounding the nucleus. If it is allowed to revolve in 2s orbit its total energy content would be zero as the helium nucleus with two 1s electrons (helium atom) is treated as electrically neutral by the approaching third electron. That is the third electron in helium atom practically has no binding energy at all. However atoms make molecules by making bond between them, which confirms the approaching atom (electron + nucleus) has some binding energy to hold neighboring atom.