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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 |
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Answer» Solution :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. |
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