Physics And Measurement Question 209
The momentum of an electron in an orbit is $ h/\lambda $ where h is a constant and lambda is wavelength associated with it. The nuclear magneton of electron of charge e and mass $ m _{e} $ is given as $ {\mu _{n}} $ $ =\frac{e h}{4\pi m _{e}} $ .The dimensions of $ {\mu _{n}} $ are $ (A\to current) $
Options:
A) $ [ML^{2}A] $
B) $ [ML^{3}A] $
C) $ [L^{2}A] $
D) $ [ML^{2}] $
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Answer:
Correct Answer: C
Solution:
$ {\mu _{n}}=\frac{ATML^{2}T^{-1}}{M}=L^{2}A $
BETA
