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In hydrogen atom the electron is making $6.610^{15}$ rev/s around the nucleus in an orbit of radius 0.528 $\dot{A}$ . The magnetic moment $\left ( A-m^{2} \right )$ will be
Option: 1
$110^{-15}$

Option: 2
$110^{-10}$

Option: 3
$110^{-23}$

Option: 4
$1*10^{-27}$

Answers (1)

best_answer

As we learnt

Maximum area -

Maximum magnetic moment

-

$i=qv$

$=1.6*10^{-19}*6.6*10^{15}$

$=10.5*10^{-4}A$

Now, Area $A=\pi R^{2}$

$=3.14*\left ( .528 \right )^{2}*10^{-20}m^{2}$

magnetic moment M=iA

$=10.5*10^{-4}*3.14*\left ( 0.528 \right )^{2}*10^{-20}m^{2}$

$=10*10^{-24}=1*10^{-23}A-m^{2}$

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rishi.raj

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