The Bohr model for the H-atom relies on the Coulomb's law of electrostatics . Coulomb's law 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`

Options

A. `-11.4`
B. `-17.3`
C. `5.9`
D. `-23.2`

Answer

A

Explanation

Given `epsilon=0.1,R_0=1Å`, Let `epsilon=2+delta`
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`.