Finding period of revolution of an electron (fairly )?

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SUMMARY

The discussion focuses on calculating the period of revolution of a doubly ionized atom with a mass of 3.65E-26 kg, accelerated by a voltage of 3950.0 V in a magnetic field of 0.0500 T. The correct radius of curvature for the ion's path is determined to be 0.601 m. The period of revolution is calculated using the formula T=√(4π²m²/(Q²B²)), resulting in a value of 0.0000143156 s. Participants emphasize the importance of using scientific notation and significant figures in reporting results.

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Finding period of revolution of an electron (fairly urgent)?

Homework Statement



A doubly ionized atom (charge = +2e) whose mass is 3.65E-26 kg is accelerated by a voltage of 3950.0 V and enters a region where a uniform magnetic field B = 0.0500 T acts perpendicular to its motion.

a) What is the radius of curvature of the path of the ion in the B-field?

b) What is its period of revolution?

Homework Equations



F=QvB
F=m4pi2r/T2
F=mv2/r

The Attempt at a Solution



I did the first part, the correct answer is 0.601 m.

I've tried a few different methods for b), I won't list them all.

I rearranged the third equation for v, plugged it into the first equation, plugged that into the second equation and rearranged for T.

T=√(4π2m2/(Q2B2))

Solved that for 0.0000143156 s

No idea if this is right or not, but I doubt it because I think I need to use voltages and energies, but that didn't work when I tried it.
 
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Your result looks okay for T, but you should use scientific notation to avoid all those leading zeros! Also trim for reasonable significant figures :smile:

If you already had the radius of the circle and speed, you could simply have found the time to travel one circumference :wink:
 


Yea, I tried that, but kept getting the wrong answer =/ I think the value I listed was right after all, thanks!
 

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