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The inverse square law in electrostatics is |F|=(e^(2))/((4pi in_(0)r^(2)) for the force between an electron and 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 photon had a mass m_(p), force would be modified to|F|=(e^(2))/((4pi in_(0))) [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 :Here, we suppose MASS of photon, `m_(p)=10^(-6)` electronic mass i.e., `m_(p)=10^(-6)(0.5)MeV=5xx10^(-7)xx1.6xx10^(-13)J=0.8xx10^(-19)J`. Now, `1/(lambda)=h/(m_(p)c)=(hc)/(m_(p)c^(2))=(10^(-34)xx3xx10^(8))/(0.8xx10^(-19))=4xx10^(-7)m` which is much larger than Bohr radius. As `|vecF|=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r] exp. (-LAMBDAR).....(i) where lambda^(-1)=h/(m_(p)c)=4xx10^(-7)mgt gt r_(B)` `:. lambdalt lt1/(r_(B)) or lambda_(B)lt lt1` Now, `U(r)=(-e^(2))/(4pi in_(0)) (exp (-lambdar))/r......(II)` As `mvr=h. v=h/(mr). As exp (-lambdar)to1`, therefor, form (i) ` (mv^(2))/r=F=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]` `:. m/r (h/(mr))^(2)=(e^(2))/(4pi in_(0))[1/(r^(2))+lambda/r]` `(h^(2))/m=(e^(2))/(4pi in_(0)) r+lambdar^(2)........(iii)` If `lambda=0, r_(B), then (h^(2))/m=(e^(2))/(4pi in_(0))r_(B)`. As `lambda^(-1)gt gt r_(B)`, put `r=r_(B)+delta` form (iii), `r_(B)=(r_(B)+delta)+lambda(r_(B)+delta)^(2)=r_(B)+delta+lambda(r_(B)^(2)+delta^(2)2delta r_(B))` Neglecting `delta^(2)`, we get `0=lambda r_(B)^(2)+delta(1+2lambda r_(B)^(2))` `delta=(-lambdar_(B)^(2))/(1+2lambdar_(B))=-lambdar_(B)^(2)(1-2lambda r_(B))` `delta=-lambdar_(B)^(2)(As lambdar_(B)lt lt1)` form (ii), `V(r)=(-e^(2))/(4pi in_(0)) (exp. (-lambda delta-lambdar_(B)))/(r_(B)+delta)=(-e^(2))/(4pi in_(0)) 1/(r_(B))[(1-delta/(r_(B))) (1-lambdar_(B))]=-27.2eV` i.e., V(r) remai ns unchanged. `K.E.=1/2mv^(2)=1/2m(h/(mr))^(2)=(h^(2))/(2m(r_(B)+delta)^(2)) = (h^(2))/(2mr_(B)^(2)) (1-(2delta)/(r_(B)))=(13.6eV)[1+2lambdar_(B)]` Total energy `=(-e^(2))/(4pi in_(0)r_(B))+(h^(2))/(2mr_(B)^(2)) (1+2lambda r_(B))=-27.2+13.6(1+2lambdar_(B))eV.` Change in ground state energy of hydrogen atom `=13.6xx2lambda_(B)eV=27.2lambdar_(B)eV` |
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