Fig. 22.43 of our textbook for a diagram of a mass spectrometer). Recall that th
ID: 2032760 • Letter: F
Question
Fig. 22.43 of our textbook for a diagram of a mass spectrometer). Recall that this device consists of two Two stages: the first is called the velocity selector and contains both an electric and a magnetic field, while the second stage is a chamber with only a magnetic field. For our analysis, suppose that in the velocity selector we use electric and magnetic fields of magnitude 207 V/m and 0.00023 T, respectively. In the second stage, we use a magnetic field of magnitude 0.23 T. See the hint below for values of constants. isotopes of an unknown element are analyzed through a mass spectrometer (refer to If one of the isotopes strikes the detector wall a distance of 0.68 m from the point where it enters the second stage, what is its mass? Number kg If the other isotope strikes the detector a distance of 0.84 m from the point where it enters the second stage estimate how many more neutrons it has compared to the first isotope. Recall that isotopes are variants the same element that have different number of neutrons. Enter your answer as a whole number Number extra neutronsExplanation / Answer
let v is the velocity of the charged particle in velocity selector.
In velocity selcector, FB = Fe
q*v*B = q*E
v = E/B
= 207/0.00023
= 9*10^5 m/s
given radius of path, r = d/2 = 0.68/2 = 0.34 m
in the second stage, B = 0.23 T
we know, r = m*v/(B*q)
m = B*q*r/v
= 0.23*1.6*10^-19*0.34/(9*10^5)
= 1.39*10^-26 kg <<<<<<-------Answer
b) let m1 = 1.39*10^-26 kg
r2 = 0.84/2 = 0.42 m
m2 = B*q*r2/v
= 0.23*1.6*10^-19*0.42/(9*10^5)
= 1.717*10^-26 kg
no of extra newutrons = (m2 - m1)/m_nuetron
= (1.717 - 1.39)*10^-26/(1.67*10^-27)
= 2 <<<<<<------------------Answer
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