How will you show that radius of first Bohr orbit (a0) is 0.529Ao with the help of permittivity of free space instead of coloumb' s law constant ?    ε0=8.854×10−12C2/Nm2,h=6.626×10−34Js.

We know, a0=ε0h2πme2 where m and e are the mass and charge on the electron.            Hence : a0=8.854×10−12C2N.m2                         ×6.626×10−34Js23.14×9.109×10−31kg×1.602×10−19C2=5.29×10−11J2s2Nm2×kg=5.29×10−11J2s2Nm×m×kg=5.29×10−11J2s2J×m×kg=5.29×10−11Js2m×kg=5.29×10−11kgm2s−2×s2m×kg                          =5.29 x 10-11m                          =5.29×10−11m×1010Ao1m=0.529Ao∵1Nm=J;J=kgm2s−2