Solving Mass Spectrometer Homework Problem with 90SR+ Isotope

In summary, the problem involves finding the voltage difference required to accelerate an isotope to a specific speed. The equations used are U=qV and U=0.5mv^2, with given values of q (1.602*10^-19 C) and m (1.495*10^-25 Kg). After setting the two equations equal to each other and solving for the kinetic energy, the answer of 30866.73 V was obtained. However, the professor's answer of approximately 308 volts is more reasonable. Upon reviewing the problem, the error was found and corrected.
  • #1
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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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  • #2
False alarm everybody. Typing out the problem helped me catch the error. Cheers!
 

Related to Solving Mass Spectrometer Homework Problem with 90SR+ Isotope

What is a mass spectrometer and how does it work?

A mass spectrometer is a scientific instrument used to determine the molecular weight and chemical composition of a substance. It works by ionizing a sample and then separating the ions based on their mass-to-charge ratio. This separation is achieved by passing the ions through an electric or magnetic field, with heavier ions being deflected less than lighter ions.

Why is the 90SR+ isotope important in mass spectrometry?

The 90SR+ isotope, also known as strontium-90, is important in mass spectrometry because it is a radioactive isotope that is commonly used for radiometric dating. This means that it can be used to determine the age of rocks and other materials, making it a valuable tool in geological and environmental research.

What are the main steps involved in solving a mass spectrometer homework problem?

The main steps involved in solving a mass spectrometer homework problem include understanding the principles of mass spectrometry, being familiar with the specific instrument being used, preparing and analyzing the sample, interpreting the data, and drawing conclusions based on the results.

How does the 90SR+ isotope affect the results of a mass spectrometer analysis?

The presence of the 90SR+ isotope in a sample will result in a peak on the mass spectrometer graph at a specific mass-to-charge ratio. This peak can be used to identify the presence of strontium-90 in the sample and determine its abundance. The 90SR+ isotope can also be used as a reference point for calibrating the mass spectrometer.

What are some common sources of error in mass spectrometer analysis?

Some common sources of error in mass spectrometer analysis include contamination of the sample, instrument malfunction or misalignment, human error in sample preparation or data interpretation, and interference from other substances in the sample. It is important to carefully control for these factors in order to obtain accurate and reliable results.

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