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Two isotopes of carbon, carbon-12 and carbon-13, have masses of 1.993 10-26 kg and 2.159 10-26 kg, respectively. These two isotopes are singly ionized ( e) and each is given a speed of 6.50 105 m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.5700 T. Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

User Vihkat
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Answer:

d=0.024m

Step-by-step explanation:

The mass of the isotopes are


m_(c12)=1.993*10^(-26)kg


m_(c13)=2.159*10^(-26)kg

The velocity of isotopes are
v=6.50*10^(5)m/s

Charge q=+e=1.6×10⁻¹⁹C and the magnetic field is B=0.5700T

The radius of circular paths for isotopes are:


r_(1)=(m_(c12)*v)/(|q|B) \\r_(1)=(1.993*10^(-26)kg*6.50*10^(5)m/s)/(1.6*10^(-19)C*0.5700T)\\ r_(1)=0.142m


r_(2)=(m_(c13)*v)/(|q|B) \\r_(2)=(2.159*10^(-26)kg*6.50*10^(5)m/s)/(1.6*10^(-19)C*0.5700T)\\ r_(2)=0.154m

So the spatial separation between two isotopes given as:

d=2r₂-2r₁

d=2(0.154)-2(0.142)

d=0.024m

User Banex
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