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in a mass spectrometer, germanium atoms have radii of curvature equal t…

Question

in a mass spectrometer, germanium atoms have radii of curvature equal to 21.0, 21.6, 21.9, 22.2, and 22.8 cm. the largest radius corresponds to an atomic mass of 76 u
part a
what are the atomic masses of the other isotopes?
express your answers using two significant figures separated by commas.

Explanation:

Step1: Recall the formula for radius in mass spectrometer

In a mass - spectrometer, for ions moving in a circular path under a magnetic field \( r=\frac{mv}{qB} \). Assuming \( v\), \(q\), and \(B\) are constant (same charge and velocity, same magnetic field for isotopes of the same element), \(r\propto m\) (mass). So, \(\frac{m_1}{m_2}=\frac{r_1}{r_2}\)

Step2: Calculate the atomic masses

Let \(m_{22.8} = 76\space u\) (given)

  • For \(r = 21.0\space cm\):

\(m_{21.0}=\frac{21.0}{22.8}\times76\space u\)
\(m_{21.0}=\frac{21.0\times76}{22.8}\space u=\frac{1596}{22.8}\space u = 70\space u\)

  • For \(r = 21.6\space cm\):

\(m_{21.6}=\frac{21.6}{22.8}\times76\space u\)
\(m_{21.6}=\frac{21.6\times76}{22.8}\space u=\frac{1641.6}{22.8}\space u = 72\space u\)

  • For \(r = 21.9\space cm\):

\(m_{21.9}=\frac{21.9}{22.8}\times76\space u\)
\(m_{21.9}=\frac{21.9\times76}{22.8}\space u=\frac{1664.4}{22.8}\space u = 73\space u\)

  • For \(r = 22.2\space cm\):

\(m_{22.2}=\frac{22.2}{22.8}\times76\space u\)
\(m_{22.2}=\frac{22.2\times76}{22.8}\space u=\frac{1687.2}{22.8}\space u = 74\space u\)

Answer:

\(70,72,73,74\)