If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1Å, and (ii) R = 10 Å.
(i) Consider in H atom nucleus as point charge electron is revolving around nucleus with speed v and radius rA. The Coulombian force provides centripetal force to revolve around nucleus. ∴rAmeν=rA2−Ke2 ...(I) Here K=4πε01 (−) sign shows the force of attraction. By Bohr's postulate, angular momentum =2πnh mνrA=2πnh ν=2πmrAnh 4π2m2rA2rAmn2h2=rA2Ke2 [From I] rA=4π2mKe2n2h2..(II) For ground state n−1 rA=4π2mKe2h2 =(2×3.14)29.1×10−31×9×109×1.6×10−19×1.6×10−196.63×10−346.63×10−34 9.19×1.6×1.6×4×3.14×4×3.146.63×6.63×10−68+38+31−9 =rA=0.53Ao P.E.=rA−Ke2=0.53×10−10−9×109×1.6×10−19×1.6×10−19J =0.53×10−10×1.6×10−19−9×1.6×1.6×10−19×10−19×109J P.E.=0.5314.4−27.17−27.2eV K.E=21mν2 =214π2m2r2m⋅n2h2=214π2mr2h2 (n = 1 for ground state) =214×3.14×3.14×9×10−31×0.53×10−10×0.53×10−106.62×10−34×6.62×10−34J =4×3.14×3.14×18×0.53×0.53×1.6×10−196.62×6.62×10−58+31+10+10eV =4×3.14×3.14×18×0.53×0.536.62×6.62×10−58+51+19 K.E=0.1373×102eV=13.7eV P.E.=27.2eV (ii) Now for spherical nucleus of radius, R , electron moves charge inside the nucleus R>>rb then electron moves inside the nucleus. Then (rb is radius of new Bohr's orbit of revolving electron) Charge=44πR3e⋅(34πrb3) e′=q2=R3er33 q1=e r3mν2=r32Kee′ (By Coulomb's law) mνr3=2πnh⇒2πmr3nh (By Bohr's postulate) ∴rbm4π2m2r32n2h2=rb2kee′ rb=4π2mkee′n2h2 Now for ground state of H, m=1 and e′=R3erA3, then ∴rb=4π2mK⋅e⋅e⋅R3rb3h2=[4π2mKe2h2]×rb3R3=rA×rb3R3 [∵rA=4π2mKe3h2=0.53Aocalclatedinpart(i)] rb=rA[rbR]3 rb4=rAR3=0.53Ao×(10Ao)3 rbA=0.53×1000(Ao)4(∵rA=0.53Ao) rb=[530]41Ao<RAo K.E.=21mν2=2m⋅4π2m2rb2h2=8π2mrb2h2[∵ν=4π2m2rb2n2h2] =8×3.14×3.14×9.1×10−31×4.8×4.8×10−20×1.6×10−196.62×6.62×10−34×10−34eV =8×3.14×3.14×9.1×4.8×4.8×1.66.62×6.62×10−68+31+20+19eV =26460.243.824410−58+70−0.001656×102eV K.E=0.167eV P.E.=4πε0e2R3(rb2−3R2)[P.E=rKq1q2] P.E.=[4πεorAe2]R3rA(rb2−3R2)[multiplyingbyrArA] From part (i) P.E.=4πεorAe2=27.2eV ∴P.E=27.2[10000.53(530−300)]⎣⎢⎡∵rb=(530)41AoandR=10Ao⎦⎥⎤ =1000A327.2eV×0.53(23.02−300)A3=27.2×10000.53(−276.9)eV P.E.=10003992.9=−3.99eV K.E.=0.167eV
