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=0.1 Å and (ii) R=10 Å`

Explanation

Orbital radius of electron in the ground state in H-atom = Bohr radius
`a_(0)=r_(1)=0.53Å=(epsi_(0)h^(2))/(pi me^(2)) ( :.n=1)...(1)`
Total energy of electro in above case,
`E_(1)=-13.6eV ....(2)`
`rArr K_(1)=-E_(1)=13.6eV ....(3)`
and `U_(1)=2E_(1)=2(-13.6)=-27.2eV ....(4)`
(Because for an electron, making orbital motion around the nucleus in an atom, `E = -K=(U)/(2))`
Here as per the statement in the first case,
`R=0.1Å rArr R lt a_(0)` ( `:.` From equatio `(1) a_(0)=0.53Å)`
`rArr` We can take proton as a point charge. Hence total energy of H-atom in this case will be
-13.6eV as per equation (2).
As per statement in the second case,
`R=10 Å " "..(5)`
`rArr R gt gt a_(0)`
`rArr` Electron would make its orbital motion inside proton. In this case, new Bohr radius is suppose `b_(0)`.
Now charge e is uniformly distributed in its volume `(4)/(3)piR^(3)`. Hence in volume `(4)/(3)pib_(0)^(3)`. if charge contained is e. then since volume density of electric charge is constant.
`(e)/((4)/(3)pi R^(3))=(e.)/((4)/(3)pi b_(0)^(3))`
`:. e.=((b_(0)^(3))/(R^(3)))e....(6)`
Now writing `b_(0)` in place of `r_(1)` and writing
`e^(2)=e.e`,
`b_(0)=(epsi_(0)h^(2))/(pi me.e)=(epsi_(0)h^(2))/(pi me)xx(1)/(e.)`
`:. b_(0)=(epsi_(0)h^(2))/(pi me)xx(R^(3))/(b_(0)^(3)e)` [ From equation (6)]
`:.b_(0)^(4)=((epsi_(0)h^(2))/(pi me^(2)))R^(3)`
`=(0.53)(10)^(3)` [From equation (1) and (5)]
`=510 Åxx(Å)^(3)`
`=510(Å)^(4)`
`:.b_(0)=(510)^(1//4)={(510)^(1//2)}^(1//2)=4.75Å....(7)`
`rArr b_(0) lt R [ :.` Here `R=10 Å) ....(8)`
Now according to formula `r_(n)=(epsi_(0)n^(2)h^(2))/(pi mZe^(2))....(9)`
and `v_(n)=(Ze^(2))/(2epsi_(0)nh) ....(10)`
Taking multiplication of above two equations,
`v_(n)r_(n)=(Ze^(2))/(2epsi_(0)nh)xx(epsi_(0)n^(2)h^(2))/(pi m Ze^(2))`
`:.v_(n)r_(n)=(nh)/(2pi m)`
`:. v_(1)r_(1)=(h)/(2pi m)`= constant ....(11)
`:.v_(1).=v_(1)((a_(0))/(b_(0))) ....(12)`
`rArr` Here, new value of kinetic energy of electron,
`K_(1).=(1)/(2)mv_(1)^(2)`
`:. K_(1).=(1)/(2)mv_(1)^(2)((a_(0))/(b_(0)))^(2)`
`=K_(1)((a_(0))/(b_(0)))^(2)`
`=13.6((0.53)/(4.75))^(2)`
`:. K_(1).=0.1693eV ......(13)`
`rArr` Now in the present case, electric potential for `r gt R` is
`V=(kQ)/(2R^(3))(3R^(2)r^(2))`
( where k= Coulomb.s constant `=(1)/(4pi epsi_(0))`
Taking V=V. and Q=e= charge of proton for `r=b_(0)`
`V.=(Ke)/(2R^(3))(3R^(2)-b_(0)^(2))`
Now new electrostatic potential energy in ground state,
`U_(1).=V_(q).`
`:.U_(1).=(ke)/(2R^(3))(3R^(2)-b_(0)^(2))(-e)` ( `:.` Charge of electron q=-e)
`=-(ke^(2))/(2R^(3))(3R(-2)-b_(0)^(2))`
`=-(9xx10^(9)xx(1.6xx10^(-19))^(2))/(2xx(10xx10^(-10))^(3)) xx{3(10xx10^(-10))^(2)-(4.75xx10^(-10))^(2)`
`=-(9xx2.56)/(2)xx10^(9-38-20)(300-22.56)`
`=-3.196xx10^(-19)J`
`=-(3.196xx10^(-19))/(1.6xx10^(-19))eV`
`:.U_(1).=-1.9975eV ...(14)`
From equation (13) and (14)
`E_(1).=K_(1).+U_(1).`
`:.E_(1).=0.1693+(-1.9975)`
`:.E_(1).=-1.8282eV`
Note : There are mistakes in the answers of NCERT exemplar book.