Hint: The electrostatic force provides the required centripetal force.
Step 1: Find the wavelength.
For mp=10-6 timesmp=10−6 times the mass of an electron, the energy associated with it is given by-
mpc2=10-6×electron mass×c2mpc2=10−6×electron mass×c2
≃10-6×0.5 MeV≃10−6×0.5 MeV
≃10-6×0.5×1.6×10-23≃10−6×0.5×1.6×10−23
≃0.8×10-19 J≃0.8×10−19 J
The wavelength associated with it is given by:
ħmpc= ħcmpc2=10-34×3×1080.8×10-19 ħmpc= ħcmpc2=10−34×3×1080.8×10−19
≈4×10-7 m>>Bohr radius≈4×10−7 m>>Bohr radius
Step 2: Find the potential energy.
|F|=e24πε0[1r2+λr]exp(-λr)|F|=e24πε0[1r2+λr]exp(−λr)
where, λ-1= ħmpc≈4×10-7m>>rB
∴λ<<1rB i.e., λrB<<1
U(r)=-e24πε0·exp(-λr)r
mvr= ħ ∴ v= ħmr
Also mv2r=≈(e24πε0)[1r2+λr]
∴ ħ2mr3=(e24πε0)[1r2+λr]
∴ ħ2m=(e24πε0)[r+πr2]
if λ=0; r=rB= ħm·4πε0e2
ħ2m=e24πε0·r
Since, λ-1>>r, put r=rB+δ
∴rB=rB+δ+λ(r2B+δ2+2δrB); neglect δ2
or 0=λr2B+δ(1+2λrB)
δ=-λr2B1+2λrB≈λr2B(1-2λrB)=-λr2B
Since, λrB<<1
∴V(r)=-e24πε0·exp(-λδ-λrB)rB+δ
∴V(r)=-e24πε01rB[(1-δrB)·(1-λrB)]
≅(-27.2eV) remains unchanged
Step 3: Find the kinetic energy.
KE=-12mv2=12m· ħ2m2r2= ħ22m[rB+δ]2= ħ22mrB2(1+δrB)-2= ħ22mrB2(1-2δrB)
Total energy=-e24πε0rB+ħ22mr2B[1+2λrB]
=-27.2+13.6[1+2λrB]eV
Change in energy=1.36×2λrBeV=27.2λrBeV