Solving Mass Spectrometer Homework Problem with 90SR+ Isotope

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SUMMARY

The discussion focuses on solving a mass spectrometer homework problem involving the isotope 90SR+. The user correctly determined the speed of the isotope to be 25.72 x 104 m/s but initially miscalculated the voltage difference required for acceleration, arriving at 30866.73 V instead of the professor's correct answer of approximately 308 V. The user identified their error while articulating the problem, demonstrating the importance of clear problem-solving communication in physics.

PREREQUISITES
  • Understanding of kinetic energy equations (U = qV and U = 0.5mv2)
  • Knowledge of elementary charge (q = 1.602 x 10-19 C)
  • Familiarity with mass of isotopes (m = 1.495 x 10-25 Kg)
  • Basic principles of mass spectrometry
NEXT STEPS
  • Review the derivation of kinetic energy equations in physics
  • Study the principles of mass spectrometry and ion acceleration
  • Learn about the relationship between voltage and kinetic energy in charged particles
  • Explore common mistakes in calculations involving orders of magnitude
USEFUL FOR

Students studying physics, particularly those focusing on mass spectrometry, as well as educators looking for insights into common calculation errors in kinetic energy problems.

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Homework Statement



I'm having problem with a couple parts of this problem. I know how to work it (I think), but I'm consistently off by a couple orders of magnitude off the answer provided by the professor.

I've already found that the isotope 90SR+ needs to be given a speed of 25.72*10^4 m/s. This much I know is correct. From there I am asked to find the voltage difference required to accelerate the isotope to this speed.


Homework Equations



U=qV
U=.5mv^2

q (elementary charge): 1.602*10^-19 C
m (mass of isotope): 1.495*10^-25 Kg

The Attempt at a Solution



Set the two equations equal to one another. Find the kinetic energy of the isotope traveling at the needed speed. Divide this value by q (charge of a +1 ion). I get the answer 30866.73 V. Professor gets roughly 308 volts. Granted, his answer makes more sense than mine, but I cannot for the life of me find the fault in my work. Note: I calculated the mass of the isotope, but since it yielded the correct answer in a previous question, I assume it is correct. Someone tell me how the hell I'm off by two orders of magnitude?
 
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False alarm everybody. Typing out the problem helped me catch the error. Cheers!
 

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