Question
Chemistry Question on Bohr’s Model for Hydrogen Atom
The inverse square law in electrostatics isF=(4πε0)⋅r2e2 for the force between an electron and a proton. The (r1) dependence of ∣F∣ can be understood in quantum theory as being due to the fact that the ??article? F=(4πε0)r2e2[r21+rλ]⋅exp(−λr) where λ=mpc/? and ?=2πh The change in the ground state energy (eV) of a H-atom if mp were 10−6 times the mass of an electron. (rB= Bohr?? radius)
18.6λrB
−27
27.2λrB
−λrB
27.2λrB
Solution
As λ=?mpc=?cmpc2 =?c(10−6me)c2 =(1.05×10−34Js)(3×108ms−1)10−6[0.51](1.6×10−16J) =0.26×107m−1[∵mec2=0.51MeV] rB (Bohr?? radius) =0.51?=0.51×10−10m or λrB=(0.26×107m−1)(0.51×10−10m) =0.14×10−3<<1 Further, as ∣F∣=(4πε0e2)[r21+rλ]e−λr...(i) and ∣F∣=drdU, Ur=∫∣F∣dr=(4πε0e2)∫(rλe−λr+r2e−λr)dr If z=re−λr=r1(e−λr), drdz=[r1(e−λr)(−λ)+(e−λr)(−r21)] or dz=−[rλe−λr+r2e−λr]dr Thus ∫(rλe−λr+r2e−λr)dr ⇒−∫dz=−z=−re−λr =−(4πε0e2)(re−λr)...(ii) We know that, mvr=? ⇒v=mr?; and rmv2=F=(4πε0e2)(r21+rλ) [puttinge−λr≈1in eqn.(i)] Thus (rm)(m2r2?2)=(4πε0e2)(r21+rλ) or mr3?2=(4πε0e2)(r3r+λr2) or m?3=(4πε0e2)(r+λr2)...(iii) When λ=0,r=rB, and m?2=(4πε0e2)rB...(iv) From eqns. (iii) and (iv),rB=r+λr2 Let r=rB+δ so that from (iii) rB=(rB+δ)+λ(rB2+δ2+2δrB) or 0=λrB2+δ(1+2λrB)(neglecting δ2) or δ=−(1+2λrB)λrB2=(−λrB2)(1+2λrB)−1 =(−λrB2)(1−2λrB)=−λrB2(∵λrB<<1) From eqn. (ii)Ur=−(4πε0e2)(rB+δ)e−λ(rB+δ) =−(4πε0e2rB1)(1−rBδ)(1−λrB)≈−4πε0rBe2 =−27.2eV [∵e−λ(rB+δ)≈1−λ(rB+δ)=1−λrB−λδ≈1−λrB] and (rB+δ)1=rB(1+δ/rB)1 =rB1(1+rBδ)−1 =rB1(1−rBδ) Further, KE of the electron, K=21mv2 =21m(m2r2?2) =2mr2?2=2m(rB+δ)2?2 =2mrB2+(1+δrB2)?2 =(2mrB2?2)(1+rBδ)−2 =(2mrB2?2)(1−rB2δ) =(13.6)(1+2λrB)eV (as 2mrB2h2=13.6eV and δ=−λrB2) Total energy of H-atom in the ground state = final energy - initial energy =(13.6+27.2λrB)eV−(−13.6eV) =(27.2λrB)eV