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The Bohr model for the H-atom relies on the Coulomb's law of electrostatics . Coulomb'slaw has not directly been varified for very short distances of the order of angstroms. Suppos-ing Coulomb's law between two oppsite charge +q_(1),-q_(2) is modified to |vec(F)|=(q_(1)q_(2))/((4piepsilon_(0))r^(2))1/r^(2),rgeR_(0) =(q_(1)q_(2))/((4piepsilon_(0))r^(2))1/R_(0)^(2)(R_(0)/r)^(epsilon), rleR_(0) Calculate in such a case , the ground state enenergy of H-atom , if epsilon= 0.1,R_(0)=1Å |
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Answer» `-11.4` In CASE of H-atom , `(q_1q_2)/(4piepsilon_0)=(1.6xx10^(-19))^2 (9xx10^9)=2.3xx10^(-28) N m^(2)` It is given that `(1/R_0^2)(R_0/R)^epsilon=(1/R_0^2)(R_0/r)^(2+delta)=(R_0^delta)/r^(2+delta)` Thus, `F=(xR_0^delta)/r^(2+delta) " " (because X=(q_1q_2)/(4piepsilon_0))` As `F=(mv^2)/rimplies (mv^2)/r=(xR_0^delta)/(r^(2+delta) " " or " "v^2=(xR_0^delta)/(mr^(1+delta))`...(i) (i)As mvr=nħ`implies`r=`(nħ)/(mv)` Using eqn (i) r=`(nħ)/m[m/(xR_0^delta)]^(1//2) r((1+delta))/2` or `r^((1+delta)//2)=((n^2ħ^2)/(mxR_0^delta))^(1//2) THEREFORE r=((n^2ħ^2)/(mxR_0^delta))^(1//(1-delta))` For n=1, `r_1=(ħ^2/(mxR_0^delta))^(1//(1-delta))` `=[(1.05xx10^(-34))^2/((9.1xx10^(-31))(2.3xx10^(-23))(10^19)]]^(1/2.9)` `=8xx10^(-11) m=0.8Å` (ii)From `v_n=(nħ)/(mr_n)=nħ((xR_0^delta)/(n^2ħ))^(1/(1-delta))` for n=1,`v_1=ħ/(mr_1)=1.44xx10^6 m s^(-1)` (iii) K.E. =`1/2mv_1^2=9.43xx10^(-19) J =5.9 eV` (iv)P.E. of electron from `oo` to `R_0` ,`U_1=1/(4piepsilon_0)((q_1q_2)/R_0)=(-x)/R_0` P.E. of electron from `R_0` to r, `U_2=-int_(R_0)^(r)Fdr` `=xR_0^deltaint_(R_0)^r (DR)/(2+delta)=(xR_0^delta)/(r^(1+delta))|1/r^(1+delta)|_(R_0)^(r)=(-x)/((1+delta))[R_0^delta/r^(1+delta)-1/R_0]` Total P.E. of electron P.E. =`-x/(1+delta)[R_0^delta/r^(1+delta)-1/R_0+(1+delta)/R_0]` `=(-2.3xx10^(-28))/0.9[R_0^delta/r^(1+delta)-delta/R_0]=(-2.3xx10^(-28))/0.9[R_0^(-19)/r^(-0.9)-1.9/R_0]` `=(-2.3xx10^(-28))/(-0.9)[(0.8)^(0.9)/(10^(-10xx(-1.9))-1.9/10^(-10)]` `=(2.3xx10^(-28))/(0.9xx10^(-10))[(0.8)^(0.9)-1.9]=-17.3 eV` Total energy `E=KE+PE=5.9-17.3 eV=-11.4 eV`. |
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