How many revolutions would an electron make in the n=2 state before decaying?

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SUMMARY

The discussion focuses on calculating the number of revolutions an electron makes in the n=2 state of a hydrogen atom before decaying, which lasts approximately 5.79 ns. The initial calculation incorrectly used the mass of the electron as 1.6727E-27 kg instead of the correct value of 9.1E-31 kg. Using the correct mass, the electron's velocity is recalculated, leading to a revised distance traveled and ultimately the correct number of revolutions, which is determined to be 2585.5 revolutions.

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Homework Statement



A hydrogen atom exists in an excited state for typically 5.79 ns. How many revolution would an electron make in the n=2 state before decaying?



Homework Equations


L=m*v*Rn=h(bar)*n
w=v/Rn
Rn=n^2*Ao
Ao=.053 nm


The Attempt at a Solution


Rn=4*.053=.212 nm (don't know what purpose this serves)
v=2*1.0546E-34/(1.6727E-27*.212E-9)
v= 594.8 m/s
distance= v*t
d= 594.8*5.79E-9s= 3.444 E-6 m
circumfrance- 2*pi*r= 2*3.14*.212E-9m=1.332E-9 m
d/circumfrance= 2585.5 revolutions

This is wrong, can someone help me?
 
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You put a wrong value for the mass of the electron. It should be 9.1 E-31 kg.
 

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