Question

Atomic weight of single charge $ Li $ is $ 6.06\,amu $ and its energy is $ 400\,eV $ . It enter normally to a uniform magnetic field of $ 0.8 $ Tesla, then radius of circular path is:

• A. $ 8.35\,cm $
• B. $ 9.23\,cm $
• C. $ 10.5\,cm $
• D. $ 11.25\,cm $

Answer 1

Answer: (A)

Radius of circular path $ r=\frac{m\upsilon }{Bq}=\frac{\sqrt{2mE}}{Bq} $ $ =\frac{\sqrt{2\times 6.06\times 1.6\times {{10}^{-27}}\times 400\times 1.6\times {{10}^{-19}}}}{1.6\times {{10}^{-19}}\times 0.8} $ $ =8.35cm $