InterviewSolution
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InterviewSolution
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The inverse square law in electrostatic is |F|=(e^(2))/((4pi epsi_(0))*r^(2)) for the force between an electrona nd a proton. The ((1)/(r)) dependence of |F| can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass m force would be modified to |F|=(e^(2))/((4pi epsi_(0))r^(2))[(1)/(r^(2))+(lambda)/(r)] exp(-lambdar) where lambda=(m_(p)c)/(h) and h=(h)/(2pi). Estimate the change in the ground state energy of a H-atom if m_(p) were 10^(-6) times the mass of an electron |
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Answer» SOLUTION : As per the statemen, `lambda=(m_(p)c)/(h)=(2pi m_(p)c)/(h)=(2pi (10^(-6)m_(e))c)/(h)` `:. Lambda=(2xx3.14xx10^(-6)xx9.1xx10^(-31)xx3xx10^(8))/(6.625xx10^(-34))` `:. Lambda=2.588xx10^(6)m^(-1)...(1)` Note here `lambda=(m_(p)c)/(h)` is not wavelength but it is only constant. (Because its unit in `m^(-1))`. Now Bohar radius `r_(B)=0.53Å=5.3xx10^(-11)m` `:. lambda gt gt r_(B) ...(2)` ACCORDING to statement, `|vecF|=(e^(2))/(4pi epsi_(0)) ((1)/(r^(2))+(lambda)/(r))e^(-lambdar) ....(3)` Since `vecF` is a CONSERVATIVE force, `|vecF|=(dU)/(dr)` `:.dU=|vecF|dr` `:.U= int|vecF|dr= int (e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))e^(-lambdar)dr ....(4)` Suppose `x=(e^(-lambdar))/(r)=(1)/(r)xxe^(-lambdar)` `:.(dx)/(dr)=(1)/(r) (e^(-lambdar)) (-lambda)+e^(lambda-r)(-(1)/(r^(2)))` `:.dx=(-(lambda)/(r)e^(-lambdar)-(e^(-lambdar))/(r^(2))dr` `:.dx=-((lambda^(-lambdar))/(r)+(e^(-lambdar))/(r^(2)))dr` `rArr int (lambdae^(lambda-r))/(r)+(e^(-lambdar))/(r^(2))-dr=- intdx=-x=-(e^(-lambdar))/(r)....(5)` From equation (4) and (5), `U=(e^(2))/(4pi epsi_(0))(-(e^(-lambdar))/(r))` `:.U-(e^(2))/(4pi epsi_(0))((e^(-lambdar))/(r))....(6)` Now, according to Bohr.s first postulate, for n=1 `mvr=(h)/(2pi)` `:.v=(h)/(2pi mr)....(7)` Here `F=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))e^(-lambdar)` `:.(MV^(2))/(r)=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r)) ( :. e^(-lambdar)~~1)` `:. (m)/(r)xx(h^(2))/(4pi^(2)m^(2)r^(2))=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))` [ From equation (7)] `:. (h^(2))/(4pi^(2)mr^(3))=(e^(2))/(4pi epsi_(0))((1)/(r^(2))+(lambda)/(r))` `:.(h^(2))/(4pi^(2)m)=(e^(2))/(4pi epsi_(0))(r+lambdar^(2)) ...(8)` Since no radiation is emitted for `r=r_(B)`, taking `lambda=0` from above equation `m_(p)=0` `(h^(2))/(4pi^(2)m)=(e^(2))/(4pi epsi_(0))(r_(B)) .....(9)` Comparing equation (8) and (9), `r_(B)=r+lambda r^(2).....(10)` Suppose `r=r_(B)+dr ....(11)`(where dr is diffrentially small) Now, `(e^(-lambdar))/(r)=(e^(-lambda(r_(B)+dr)))/(r_(B)+dr)` `=(1)/(r_(B)+dr){1-lambda (r_(B)+dr)}` `=(1-lambdar_(B)-lambdadr)/(r_(B)+dr)` `~~(1-lambda r_(B))/(r_(B)(1+(dr)/(r_(B)))) ( :. lambda dr` is negligible) `:.(e^(-lambdar))/(r)=((1-lambdar_(B))/(r_(B)))(1+(dr)/(r_(B)))^(-1)` `=((1-lambdar_(B))/(r_(B)))(1-(dr)/(r_(B)))` `=(1-lambdar_(B))((1)/(r_(B))-(dr)/(r_(B)^(2)))` `=(1-0)((1)/(r_(B))-0)` `( :. lambdar_(B) lt lt TL lt 1 and (dr)/(r_(B)^(2))` is negligible) `=(1)/(r_(B)) ....(12)` From equation (6) and (12), `U=-(e^(2))/(4pi epsi_(0))((1)/(r_(B)))` `=-(ke^(2))/(r_(B))( :. k=(1)/(4 pi epsi_(0)))` `=-((9xx10^(9))(1.6xx10^(-19))^(2))/((0.53xx10^(-10))` `=-4.374xx10^(10-38+10)J` `=-4.374xx10^(-18)J` `:.U=-(4.374xx10^(-18))/(1.6xx10^(-19))eV` `=-2.717xx10^(1)eV` `:.U=-27.17eV` In the present, case, kinetic energy of electron `K=(1)/(2)mv^(2)=(1)/(2)m((h^(2))/(4pi^(2)m^(2)r^(2)))` [ from equ. (7)] `:.K= (h^(2))/(8 pi^(2)mr^(2))` `=(h^(2))/(8pi^(2)m(r_(B)+dr)^(2))` `=(h^(2))/(8pi^(2)m{r_(B)(1+(dr)/(r_(B)))}^(2))` `=(h^(2))/(8pi^(2)mr_(B)^(2))(1+(dr)/(r_(B)))^(-2)` `:.K=(h^(2))/(8 pi^(2)mr_(B)^(2))(1-(2dr)/(r_(B)))....(14)` (According to binomial theorem) From equation `(10),r_(B)=r+lambdar^(2)` But according to equation (11), `r=r_(B)+dr` and so `r_(B)=r_(B)+dr+lambda(r_(B)+dr)^(2)` `:.0=dr+lambdar_(B)^(2)+2lambdar_(B)dr+lambda dr^(2)` `:. 0~~ dr+lambdar_(B)^(2)` `( :.2lambdar_(B)dr and lambda dr^(2)` are negligible here) `:.dr =-lambda_(B)^(2)....(15)` From equation (14) and (15) `K=(h^(2))/(8pi^(2)mr_(B)^(2)){1-(2)/(r_(B))(-lambda_(B)^(2))}` `:.K=(h^(2))/(8pi^(2)mr_(B)^(2))(1+2lambdar_(B))` `:.K=13.6(1+2lambdar_(B))....(16)` `( :. (h^(2))/(8pi^(2)mr_(B)^(2))=13.6eV)` Adding equation (13) and (16), total energyin final condition, `E_(f)=13.6(1+2lambdar_(B))-27.2eV` but `E_(i)=-13.6eV` Required change of total energy, `DeltaE=E_(f)-E_(i)` `=13.6(1+2lambdar_(B))-27.2-(-13.6)` `=13.6(1+2lambdar_(B))-13.6` `=13.6+27.2lambdar_(B)-13.6` `:.DeltaE=27.2lambdar_(B)eV` |
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