A) \[8.99\times {{10}^{-24}}A\,{{m}^{2}}\]
B) \[9.27\times {{10}^{-24}}A\,{{m}^{2}}\]
C) \[5.66\times {{10}^{-24}}A\,{{m}^{2}}\]
D) \[9.27\times {{10}^{-28}}A\,{{m}^{2}}\]
Correct Answer: B
Solution :
(b) \[9.27\times {{10}^{-24}}A\,{{m}^{2}}\] Bohr Magneton \[{{({{\mu }_{l}})}_{\min }}={{\mu }_{B}}=\frac{e}{4\pi {{m}_{e}}}h\] Here, \[e=1.6\times {{10}^{-19}}\text{ }C,\text{ }h=6.64\times {{10}^{-34}}Js\] \[{{m}_{e}}=9.1\times {{10}^{-13}}kg\] \[\therefore {{\mu }_{B}}\frac{160\times {{10}^{-19}}\times 6.64\times {{10}^{-34}}}{4\times 3.14\times 9.1\times {{10}^{-31}}}\] \[=9.27\times {{10}^{-24}}\,A\,{{m}^{2}}\]You need to login to perform this action.
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