Q.
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.
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answer is 0.529.
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Detailed Solution
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
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