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Units And Measurement

Question
CBSEENPH11017697

An electron of mass 9.1 x 10-31 kg revolves in a circle of radius 0.529 Å around the nucleus with speed of 2.2x106 m/s. Show that its angular momentum is equal to h/2 π, where h is Planck's constant.

Solution

Mass of the electron = 9.1 cross times 10-31 kg
Radius of the circle, r = 0.529 straight A with 0 on top 
Speed = 2.2 x 106 m/s
The angular momentum of electron is,
        L = mvr
           = 9.1 x 10-31 x 2.19 x 10x 0.527 x 10-10 

           = 1.0502 x 10-34 J s
Also space space space straight h space equals space fraction numerator straight h over denominator 2 straight pi end fraction equals fraction numerator 6.6 cross times 10 to the power of negative 34 end exponent over denominator 2 cross times 3.142 end fraction equals 1.0502 cross times 10 to the power of negative 34 end exponent space Js
Thus,  L =fraction numerator straight h over denominator 2 straight pi end fraction