Question

An electron revolving in an orbit of radius 0.5 A in a hydrogen atom executes $ {{10}^{16}} $ rev / s. The magnetic moment of electron due to its orbital motion will be

• A. $ 1256\times {{10}^{-26}}\,A-{{m}^{2}} $
• B. $ 6.53\times {{10}^{-26}}\,A-{{m}^{2}} $
• C. zero
• D. $ 256\times {{10}^{-26}}\,A-{{m}^{2}} $

Answer 1

Answer: (A)

$ M=I\,A=(eV)\,\pi {{r}^{2}} $ $ =1.6\times {{10}^{-19}}\times {{10}^{16}}\times \frac{22}{7}\times {{(0.5\times {{10}^{-10}})}^{2}} $ $ =1256\times {{10}^{-26}}A{{m}^{2}} $