The image shown represents a mass spectrometer, which is used to identify the va
ID: 1549197 • Letter: T
Question
The image shown represents a mass spectrometer, which is used to identify the various molecules within a sample, by measuring their charge-to-mass ratio. The sample is ionized and positive ions are accelerated from rest through a potential difference delta V, where they enter a region of uniform magnetic field (B= 0.248 T, pointing out of the page). The ionized molecules travel in circular trajectories, which will pass through a hole to the Detector if the trajectory has the correct size (d=5, 6 cm). The accompanying table lists atomic masses in Daltons(1 Da=1 u). Using this information, determine the accelerating potential, delta v, which corresponds to the detection of the following ions: N^+_2 O^2+_2 CO^+ If you wanted to detect cyanide, CN^-, what changes would you need to make to the setup of the spectrometer?Explanation / Answer
let m is the mass and q is the charge of the particle.
let delta_V is the potential diffrence.
workdone on charged particle = gain in kinetic energy of the paricle.
q*delta_V = (1/2)*m*v^2
==> v = sqrt(2*q*delta_V/m)
we know, radius of orbit, r = m*v/(B*q)
d = 2*m*v/(B*q)
d = 2*m*sqrt(2*q*delta_V/m)/(B*q)
d^2 = 2*m^2*2*q*delta_V/(m*B^2*q^2)
delta_V = B^2*d^2*q/(4*m)
a) N2+
delta_V = B^2*d^2*q/(4*m)
= 0.248^2*0.056^2*1.6*10^-19/(4*24*1.66*10^-27)
= 193.6 volts
b) O2 2+
delta_V = B^2*d^2*q/(4*m)
= 0.248^2*0.056^2*2*1.6*10^-19/(4*2*16*1.66*10^-27)
= 290.5 volts
c) CO+
delta_V = B^2*d^2*q/(4*m)
= 0.248^2*0.056^2*1.6*10^-19/(4*28*1.66*10^-27)
= 166 volts
d) we need to change the direction of B. B must be into the page.
or if we keep the direction of B as swon in fig, we need to move detector towards up.
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