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Chemistry Question on Bohr’s Model for Hydrogen Atom

The inverse square law in electrostatics isF=e2(4πε0)r2\left|\vec{F}\right| = \frac{e^{2}}{\left(4\pi\varepsilon_{0}\right)\cdot r^{2}} for the force between an electron and a proton. The (1r)(\frac{1}{r}) dependence of F|\vec{F}| can be understood in quantum theory as being due to the fact that the ??article? F=e2(4πε0)r2[1r2+λr]exp(λr)\left|\vec{F}\right| = \frac{e^{2}}{\left(4\pi\varepsilon_{0}\right)r^{2}}\left[\frac{1}{r^{2}} + \frac{\lambda}{r}\right] \cdot exp \left(-\lambda r\right) where λ=mpc/?\lambda = m_pc/? and ?=h2π?= \frac{h}{2\pi} The change in the ground state energy (eV)(eV) of a HH-atom if mpm_p were 10610^{-6} times the mass of an electron. (rB=r_B = Bohr?? radius)

A

18.6λrB18.6\, \lambda r_B

B

27-27

C

27.2λrB27.2 \, \lambda r_B

D

λrB-\lambda r_B

Answer

27.2λrB27.2 \, \lambda r_B

Explanation

Solution

As λ=mpc?=mpc2?c\lambda = \frac{m_{p}c}{?} = \frac{m_{p}c^{2}}{?c} =(106me)c2?c= \frac{\left(10^{-6}m_{e}\right)c^{2}}{?c} =106[0.51](1.6×1016J)(1.05×1034Js)(3×108ms1) =\frac{ 10^{-6}\left[0.51\right]\left(1.6\times10^{-16} J\right)}{\left(1.05\times10^{-34} J s\right)\left(3\times10^{8} ms^{-1}\right)} =0.26×107m1[mec2=0.51MeV]= 0.26 \times10^{7} m^{-1}\quad\left[\because m_{e}c^{2} = 0.51 MeV\right] rBr_{B} (Bohr?? radius) =0.51?=0.51×1010m = 0.51 ? = 0.51 \times10^{-10} m or λrB=(0.26×107m1)(0.51×1010m)\lambda r_{B} = \left(0.26 \times10^{7} m^{-1}\right)\left(0.51\times10^{-10}m\right) =0.14×103<<1= 0.14 \times10^{-3} < < 1 Further, as F=(e24πε0)[1r2+λr]eλr...(i)|F| = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left[\frac{1}{r^{2}} +\frac{\lambda}{r}\right]e^{-\lambda r}\quad...\left(i\right) and F=dUdr\left|F\right| = \frac{dU}{dr}, Ur=Fdr=(e24πε0)(λeλrr+eλrr2)drU_{r} = \int\left|F\right|dr = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\int\left(\frac{\lambda e^{-\lambda r}}{r} + \frac{e^{-\lambda r}}{r^{2}}\right)dr If z=eλrr=1r(eλr)z=\frac{e^{-\lambda r}}{r} = \frac{1}{r}\left(e^{-\lambda r}\right), dzdr=[1r(eλr)(λ)+(eλr)(1r2)]\frac{dz}{dr} = \left[\frac{1}{r}\left(e^{-\lambda r}\right)\left(-\lambda\right) +\left(e^{-\lambda r}\right)\left(-\frac{1}{r^{2}}\right)\right] or dz=[λeλrr+eλrr2]drdz = -\left[\frac{\lambda e^{-\lambda r}}{r} +\frac{e^{-\lambda r}}{r^{2}}\right]dr Thus (λeλrr+eλrr2)dr\int\left(\frac{\lambda e^{-\lambda r}}{r} +\frac{e^{-\lambda r}}{r^{2}}\right)dr dz=z=eλrr\Rightarrow-\int dz = -z = -\frac{e^{-\lambda r}}{r} =(e24πε0)(eλrr)...(ii)= -\left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{e^{-\lambda r}}{r}\right)\quad...\left(ii\right) We know that, mvr=?mvr = ? v=?mr;\Rightarrow v= \frac{?}{mr}; and mv2r=F=(e24πε0)(1r2+λr)\frac{mv^{2}}{r} = F = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{1}{r^{2}} +\frac{\lambda}{r}\right) [puttingeλr1in eqn.(i)]\left[\text{putting} e^{-\lambda r} \approx 1 {\text{in eqn}}.\left(i\right)\right] Thus (mr)(?2m2r2)=(e24πε0)(1r2+λr)\left(\frac{m}{r}\right)\left(\frac{?^{2}}{m^{2}r^{2}}\right) = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{1}{r^{2}} + \frac{\lambda}{r}\right) or ?2mr3=(e24πε0)(r+λr2r3)\frac{?^{2}}{mr^{3} } = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{r+\lambda r^{2}}{r^{3}}\right) or ?3m=(e24πε0)(r+λr2)...(iii)\frac{?^{3}}{m} = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(r+\lambda r^{2}\right)\quad...\left(iii\right) When λ=0,r=rB\lambda = 0, r= r_{B}, and ?2m=(e24πε0)rB...(iv)\frac{?^{2}}{m} = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)r_{B} \quad...\left(iv\right) From eqns. (iii)\left(iii\right) and (iv),rB=r+λr2\left(iv\right), r_{B} = r + \lambda r^{2} Let r=rB+δr = r_{B} + \delta so that from (iii)\left(iii\right) rB=(rB+δ)+λ(rB2+δ2+2δrB)r_{B} =\left(r_{B} +\delta\right) + \lambda\left(r_{B}^{2} +\delta^{2} +2\delta r_{B}\right) or 0=λrB2+δ(1+2λrB)0 = \lambda r_{B}^{2} + \delta\left(1+2\lambda r_{B}\right) \quad(neglecting δ2)\delta^{2}) or δ=λrB2(1+2λrB)=(λrB2)(1+2λrB)1\delta = -\frac{\lambda r_{B}^{2}}{\left(1+2\lambda r_{B}\right)} = \left(-\lambda r_{B}^{2}\right)\left(1+2\lambda r_{B}\right)^{-1} =(λrB2)(12λrB)=λrB2(λrB<<1)= \left(-\lambda r_{B}^{2}\right)\left(1-2\lambda r_{B}\right) = -\lambda r_{B}^{2} \quad\left(\because\lambda r_{B} < < 1\right) From eqn. (ii)Ur=(e24πε0)eλ(rB+δ)(rB+δ)\left(ii\right) U_{r} =-\left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) \frac{e^{-\lambda\left(r_{B}+\delta\right)}}{\left(r_{B} +\delta\right)} =(e24πε01rB)(1δrB)(1λrB)e24πε0rB = -\left(\frac{e^{2}}{4\pi\varepsilon_{0}} \frac{1}{r_{B}}\right)\left(1-\frac{\delta}{r_{B}}\right)\left(1-\lambda r_{B}\right) \approx -\frac{e^{2}}{4\pi\varepsilon_{0} r_{B}} =27.2eV= -27.2 eV [eλ(rB+δ)1λ(rB+δ)=1λrBλδ1λrB] \left[\because e^{-\lambda\left(r_{B}+\delta\right)} \approx1 -\lambda \left(r_{B} +\delta\right) = 1 -\lambda r_{B} -\lambda\delta \approx1-\lambda r_{B}\right] and 1(rB+δ)=1rB(1+δ/rB)\frac{1}{\left(r_{B} +\delta\right)} = \frac{1}{ r_{B}\left(1+\delta/ r_{B}\right)} =1rB(1+δrB)1= \frac{1}{r_{B} }\left(1+\frac{\delta}{r_{B}}\right)^{-1} =1rB(1δrB)= \frac{1}{r_{B}}\left(1-\frac{\delta}{r_{B}}\right) Further, KE of the electron, K=12mv2K= \frac{1}{2}mv^{2} =12m(?2m2r2)= \frac{1}{2}m\left(\frac{?^{2}}{m^{2}r^{2}}\right) =?22mr2=?22m(rB+δ)2= \frac{?^{2}}{2mr^{2}}= \frac{?^{2}}{2m\left(r_{B}+\delta\right)^{2}} =?22mrB2+(1+δrB2)= \frac{?^{2}}{2mr_{B}^{2} +\left(1+\delta r_{B}^{2}\right)} =(?22mrB2)(1+δrB)2 = \left(\frac{?^{2}}{2mr_{B}^{2}}\right)\left(1+\frac{\delta}{r_{B}}\right)^{-2} =(?22mrB2)(12δrB)=\left(\frac{?^{2}}{2mr_{B}^{2}}\right)\left(1-\frac{2\delta}{r_{B}}\right) =(13.6)(1+2λrB)eV= \left(13.6\right)\left(1+2\lambda r_{B}\right)eV (as h22mrB2=13.6eV\frac{h^{2}}{2mr_{B}^{2}} = 13.6 eV and δ=λrB2)\delta = -\lambda r_{B}^{2}) Total energy of HH-atom in the ground state == final energy - initial energy =(13.6+27.2λrB)eV(13.6eV)= (13.6 + 27.2 \lambda r_B) \,eV - (-13.6 \,eV) =(27.2λrB)eV= (27.2 \lambda r_B) \,eV